Chapter 2 Polynomials (Class 9 - Maths NCERT Exemplar Solutions)

Given:

The above equation is an identity which holds true for all values of $x$.

To Find:

The value of the constant $k$.

Solution:

We start by expanding the expression on the Right Hand Side (RHS) of the given equation.

Using the distributive property of multiplication (FOIL method):

$\text{RHS} = x(x + 3) + 2(x + 3)$

$\text{RHS} = x^2 + 3x + 2x + 6$

Now, substitute the expanded RHS back into equation (i):

$x^2 + kx + 6 = x^2 + 5x + 6$

Since the two polynomials are equal for all values of $x$, their corresponding coefficients must be equal.

Comparing the coefficients of the variable $x$ on both sides:

Therefore, the value of $k$ is 5.

Comparing the result with the given options, we find that the correct choice is (C).

To Find: Identify which of the given expressions is a polynomial.

Solution:

An expression is a polynomial only if the exponent of the variable in every term is a whole number ($0, 1, 2, ...$).

Evaluating the options:

(A) $\displaystyle \frac{x^2}{2} - \frac{2}{x^2} = \frac{1}{2}x^2 - 2x^{-2}$. Since $-2$ is not a whole number, it is not a polynomial.

(B) $\sqrt{2x} - 1 = \sqrt{2}x^{\frac{1}{2}} - 1$. Since $\frac{1}{2}$ is not a whole number, it is not a polynomial.

(C) Consider the expression: $x^2 + \displaystyle \frac{3x^{\frac{3}{2}}}{\sqrt{x}}$

Simplifying the second term:

So, the expression becomes $x^2 + 3x$. Here, both exponents ($2$ and $1$) are whole numbers. Thus, it is a polynomial.

(D) $\displaystyle \frac{x - 1}{x + 1}$ is a rational expression but not a polynomial because the variable is in the denominator.

The correct option is (C).

Solution:

The given number $\sqrt{2}$ is a non-zero constant.

Any non-zero constant can be written as a polynomial by multiplying it with $x^0$.

$\sqrt{2} = \sqrt{2} \cdot x^0$

The degree of a polynomial is the highest power of the variable present in the expression. In this case, the power of $x$ is $0$.

Therefore, the degree of the constant polynomial $\sqrt{2}$ is 0.

The correct option is (B).

Solution:

The degree of a polynomial is determined by the highest power of the variable whose coefficient is not zero.

Let's look at the given terms:

1. The term $0x^5$ has a coefficient of 0, so it does not exist.

2. The term $0x^3$ has a coefficient of 0, so it does not exist.

3. The remaining non-zero terms are $4x^4$, $5x$, and $7$.

Among the non-zero terms, the highest power of $x$ is 4.

The correct option is (A).

Solution:

The zero polynomial is defined as $p(x) = 0$.

Since $0$ can be written as $0x^0$, $0x^1$, $0x^{100}$, etc., there is no single highest power that can be assigned as the degree.

Therefore, by convention in mathematics, the degree of the zero polynomial is Not defined.

The correct option is (D).

Given:

$p(x) = x^2 - 2\sqrt{2}x + 1$

To Find: The value of $p(2\sqrt{2})$.

Solution:

Substitute $x = 2\sqrt{2}$ in the given polynomial:

$p(2\sqrt{2}) = (2\sqrt{2})^2 - 2\sqrt{2}(2\sqrt{2}) + 1$

Calculating the square term:

$(2\sqrt{2})^2 = 2^2 \cdot (\sqrt{2})^2 = 4 \cdot 2 = 8$

Calculating the second term:

$2\sqrt{2} \cdot 2\sqrt{2} = 4 \cdot 2 = 8$

Substitute these values back:

$p(2\sqrt{2}) = 0 + 1 = 1$

The correct option is (B).

Solution:

Let $p(x) = 5x - 4x^2 + 3$.

We need to find the value at $x = -1$:

$p(-1) = 5(-1) - 4(-1)^2 + 3$

$p(-1) = -5 - 4(1) + 3$

$p(-1) = -5 - 4 + 3$

$p(-1) = -9 + 3$

The correct option is (A).

Given:

Solution:

First, we find $p(-x)$ by replacing $x$ with $-x$ in equation (i):

Now, add $p(x)$ and $p(-x)$:

$p(x) + p(-x) = (x + 3) + (-x + 3)$

$p(x) + p(-x) = x - x + 3 + 3$

The correct option is (D).

Solution:

A zero of a polynomial $p(x)$ is a number $c$ such that $p(c) = 0$.

The zero polynomial is defined as $p(x) = 0$ for all values of $x$.

Since the value of the zero polynomial is always $0$ regardless of the input, any real number substituted for $x$ will satisfy the condition $p(x) = 0$.

Therefore, any real number is a zero of the zero polynomial.

The correct option is (C).

To Find: The zero of the polynomial $p(x) = 2x + 5$.

Solution:

To find the zero of the polynomial, we set $p(x) = 0$:

Transposing $5$ to the Right Hand Side (RHS):

$2x = -5$

Dividing both sides by $2$:

$\displaystyle x = -\frac{5}{2}$

The correct option is (B).

Solution:

Let $p(x) = 2x^2 + 7x - 4$. To find the zeroes, we factorise the quadratic polynomial using the splitting the middle term method:

$p(x) = 2x^2 + 8x - x - 4$

$p(x) = 2x(x + 4) - 1(x + 4)$

To find the zeroes, set $p(x) = 0$:

$(2x - 1)(x + 4) = 0$

This gives two possibilities:

1. $2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$

2. $x + 4 = 0 \implies x = -4$

Among the given options, $\frac{1}{2}$ is present.

The correct option is (B).

Given:

Dividend $p(x) = x^{51} + 51$

Divisor $g(x) = x + 1$

Solution:

According to the Remainder Theorem, if a polynomial $p(x)$ is divided by $(x - a)$, the remainder is $p(a)$.

Here, the divisor is $x + 1$, which can be written as $x - (-1)$. Thus, $a = -1$.

$\text{Remainder} = p(-1)$

Substituting $x = -1$ in $p(x)$:

$p(-1) = (-1)^{51} + 51$

Since the power $51$ is odd, $(-1)^{51} = -1$:

$p(-1) = 50$

The correct option is (D).

Given:

$(x + 1)$ is a factor of $p(x) = 2x^2 + kx$.

Solution:

According to the Factor Theorem, if $(x + 1)$ is a factor of $p(x)$, then $p(-1) = 0$.

Substitute $x = -1$ in $p(x)$:

$p(-1) = 2(-1)^2 + k(-1)$

$0 = 2 - k$

The correct option is (C).

Solution:

By the Factor Theorem, $(x + 1)$ is a factor of $p(x)$ if $p(-1) = 0$. Let's test the options:

For (A): $p(-1) = (-1)^3 + (-1)^2 - (-1) + 1 = -1 + 1 + 1 + 1 $$ = 2 \neq 0$

For (B): $p(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0$

Since $p(-1) = 0$ for option (B), $(x + 1)$ is a factor of $x^3 + x^2 + x + 1$.

The correct option is (B).

Solution:

Let the expression be $E = (25x^2 - 1) + (1 + 5x)^2$.

Using the identity $a^2 - b^2 = (a - b)(a + b)$ for the first term:

$25x^2 - 1 = (5x)^2 - (1)^2 = (5x - 1)(5x + 1)$

The expression becomes:

$E = (5x - 1)(5x + 1) + (5x + 1)^2$

Taking $(5x + 1)$ as a common factor:

$E = (5x + 1) [ (5x - 1) + (5x + 1) ]$

$E = (5x + 1) [ 5x - 1 + 5x + 1 ]$

The factors are $(5x + 1)$ and $10x$. Among the options, $10x$ is present.

The correct option is (D).

Solution:

To find the value of $249^2 - 248^2$, we use the algebraic identity:

Substituting $a = 249$ and $b = 248$:

$249^2 - 248^2 = (249 - 248)(249 + 248)$

$249^2 - 248^2 = (1)(249 + 248)$

Comparing the result with the given options, we find that the correct choice is (D).

Solution:

To factorise the quadratic polynomial $4x^2 + 8x + 3$, we use the method of splitting the middle term.

We need to find two numbers whose product is $4 \times 3 = 12$ and whose sum is $8$. Such numbers are $6$ and $2$.

$4x^2 + 6x + 2x + 3$

Grouping the terms:

$2x(2x + 3) + 1(2x + 3)$

The correct option is (B).

Solution:

Let the given expression be $E$. Using the expansion of $(x + y)^3$:

$(x + y)^3 = x^3 + y^3 + 3xy(x + y)$

Substituting this in the expression $E$:

$E = [x^3 + y^3 + 3xy(x + y)] - (x^3 + y^3)$

$E = x^3 + y^3 + 3xy(x + y) - x^3 - y^3$

The factors of the expression are $3$, $x$, $y$, and $(x + y)$. Since $3xy$ is a product of these factors, it is a factor of the expression.

The correct option is (D).

Solution:

We expand the binomial using the identity $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$(x + 3)^3 = x^3 + 3(x^2)(3) + 3(x)(3^2) + 3^3$

$(x + 3)^3 = x^3 + 9x^2 + 3(x)(9) + 27$

In the above expansion, the coefficient of $x$ is 27.

The correct option is (D).

Given:

Solution:

From equation (i), taking the LCM:

$\displaystyle \frac{x^2 + y^2}{xy} = -1$

$x^2 + y^2 = -xy$

We know the identity for $x^3 - y^3$ is:

$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

Substituting the value from equation (ii):

$x^3 - y^3 = (x - y) \times 0$

The correct option is (C).

Solution:

Consider the Right Hand Side (RHS) of the equation:

$\text{RHS} = \displaystyle \left( 7x + \frac{1}{2} \right) \left( 7x - \frac{1}{2} \right)$

Using the identity $(a + b)(a - b) = a^2 - b^2$:

$\text{RHS} = \displaystyle (7x)^2 - \left( \frac{1}{2} \right)^2$

Now, equate the LHS and the RHS:

$\displaystyle 49x^2 - b = 49x^2 - \frac{1}{4}$

Comparing both sides, we get:

The correct option is (C).

Solution:

We know the algebraic identity:

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

According to the given condition, $a + b + c = 0$. Substituting this into the identity:

$a^3 + b^3 + c^3 - 3abc = (0) \times (a^2 + b^2 + c^2 - ab - bc - ca)$

$a^3 + b^3 + c^3 - 3abc = 0$

The correct option is (C).

(i) $\frac{1}{\sqrt{5}}x^{\frac{1}{2}} + 1$ is a polynomial.

False.

Justification:

A polynomial in one variable $x$ is an algebraic expression where the powers of $x$ are non-negative integers (0, 1, 2, 3, ...).

In the given expression $\frac{1}{\sqrt{5}}x^{\frac{1}{2}} + 1$, the term $\frac{1}{\sqrt{5}}x^{\frac{1}{2}}$ has the variable $x$ raised to the power of $\frac{1}{2}$.

Since $\frac{1}{2}$ is a fraction and not a non-negative integer, the expression is not a polynomial.

(ii) $\frac{6\sqrt{x} \;+\; x^{\frac{3}{2}}}{\sqrt{x}}$ is a polynomial, x ≠ 0.

True.

Justification:

We can simplify the expression by rewriting the terms using exponent notation and distributing the division by $\sqrt{x}$.

$\sqrt{x} = x^{\frac{1}{2}}$

The expression is $\frac{6x^{\frac{1}{2}} \;+\; x^{\frac{3}{2}}}{x^{\frac{1}{2}}}$.

Divide each term in the numerator by the denominator:

$= \frac{6x^{\frac{1}{2}}}{x^{\frac{1}{2}}} + \frac{x^{\frac{3}{2}}}{x^{\frac{1}{2}}}$

Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

$= 6x^{\frac{1}{2} - \frac{1}{2}} + x^{\frac{3}{2} - \frac{1}{2}}$

$= 6x^0 + x^{\frac{3-1}{2}}$

$= 6x^0 + x^{\frac{2}{2}}$

$= 6x^0 + x^1$

Since $x^0 = 1$ (for $x \neq 0$ as given), the expression simplifies to:

$= 6(1) + x$

$= 6 + x$

The simplified expression is $x + 6$.

In this expression, the powers of $x$ are 1 (for the term $x$) and 0 (for the constant term 6, which can be written as $6x^0$). Both 1 and 0 are non-negative integers.

Therefore, the simplified expression is a polynomial.

An algebraic expression is a polynomial if the exponents of the variables are non-negative integers.

(i) 8

This is a polynomial.

Justification: This is a constant term. A constant can be written as $8x^0$. The exponent of the variable ($0$) is a non-negative integer. It is a constant polynomial.

(ii) $\sqrt{3}$x² – 2x

This is a polynomial.

Justification: The expression is $\sqrt{3}x^2 - 2x^1$. The exponents of the variable $x$ are 2 and 1, which are non-negative integers. The coefficients ($\sqrt{3}$ and -2) can be any real numbers.

(iii) 1 – $\sqrt{5x}$

This is not a polynomial.

Justification: The term $\sqrt{5x} = \sqrt{5} \times \sqrt{x} = \sqrt{5}x^{\frac{1}{2}}$. The expression is $1 - \sqrt{5}x^{\frac{1}{2}}$. The exponent of the variable $x$ is $\frac{1}{2}$, which is a fraction and not a non-negative integer.

(iv) $\frac{1}{5x^{-2}}$ + 5x + 7

This is a polynomial.

Justification: Simplify the first term: $\frac{1}{5x^{-2}} = \frac{1}{5} \times \frac{1}{x^{-2}} = \frac{1}{5} \times x^2$. The expression is $\frac{1}{5}x^2 + 5x + 7$. The exponents of the variable $x$ are 2 and 1 (for $5x = 5x^1$), and 0 (for the constant term $7 = 7x^0$). These exponents (2, 1, 0) are non-negative integers.

(v) $\frac{(x \;-\; 2) (x \;-\; 4)}{x}$

This is not a polynomial.

Justification: Expand the numerator: $(x-2)(x-4) = x^2 - 4x - 2x + 8 = x^2 - 6x + 8$. The expression is $\frac{x^2 - 6x + 8}{x}$. Divide each term in the numerator by $x$:

$\frac{x^2}{x} - \frac{6x}{x} + \frac{8}{x} = x - 6 + \frac{8}{x} = x - 6 + 8x^{-1}$

The term $8x^{-1}$ has the variable $x$ raised to the power of -1, which is a negative integer. Therefore, the expression is not a polynomial.

(vi) $\frac{1}{x \;+\; 1}$

This is not a polynomial.

Justification: The variable appears in the denominator in a way that cannot be simplified to terms with non-negative integer exponents. This is a rational expression, but not a polynomial.

(vii) $\frac{1}{7}$a³ – $\frac{2}{\sqrt{3}}$a² + 4a – 7

This is a polynomial.

Justification: The expression is $\frac{1}{7}a^3 - \frac{2}{\sqrt{3}}a^2 + 4a^1 - 7a^0$. The exponents of the variable $a$ are 3, 2, 1, and 0, which are non-negative integers. The coefficients ($\frac{1}{7}$, $-\frac{2}{\sqrt{3}}$, 4, -7) are real numbers.

(viii) $\frac{1}{2x}$

This is not a polynomial.

Justification: The expression can be written as $\frac{1}{2} \times \frac{1}{x} = \frac{1}{2}x^{-1}$. The exponent of the variable $x$ is -1, which is a negative integer. Therefore, the expression is not a polynomial.

The expressions that are polynomials are (i), (ii), (iv), and (vii).

(i) A binomial can have atmost two terms.

False.

Justification:

A binomial is defined as a polynomial with exactly two non-zero terms. The word "atmost" means "at most", implying a maximum of two terms. However, the definition requires exactly two terms. A polynomial with one term is a monomial, and a polynomial with three terms is a trinomial. A binomial cannot have one term (monomial) or more than two terms (like a trinomial or other polynomials).

Example of a binomial: $x + 1$, $x^2 - 5$, $3y^5 + 2y$. These have exactly two terms.

Example of a polynomial with "at most two terms": This would include monomials (e.g., $x$) and binomials (e.g., $x+1$). But a binomial itself must have two terms.

(ii) Every polynomial is a binomial.

False.

Justification:

A polynomial is a general expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Binomials are a specific type of polynomial that has exactly two non-zero terms.

Examples of polynomials that are not binomials include monomials (e.g., $5x^2$) and trinomials (e.g., $x^2 + x + 1$).

(iii) A binomial may have degree 5.

True.

Justification:

The degree of a polynomial is the highest power of the variable with a non-zero coefficient. A binomial is a polynomial with exactly two terms. The degree is determined by the term with the highest exponent.

Example of a binomial with degree 5: $x^5 + 1$, $2x^5 - 3x$. In both cases, there are exactly two terms, and the highest power of the variable is 5.

(iv) Zero of a polynomial is always 0.

False.

Justification:

A zero of a polynomial $p(x)$ is a value of $x$ for which $p(x) = 0$. This value is not necessarily 0.

Example: Consider the polynomial $p(x) = x - 5$. The zero is found by setting $p(x) = 0$, so $x - 5 = 0$, which gives $x = 5$. Here, the zero is 5, not 0.

Consider the polynomial $p(x) = 2x$. The zero is found by setting $p(x) = 0$, so $2x = 0$, which gives $x = 0$. In this case, the zero is 0. This shows that 0 can be a zero, but it's not always the zero.

(v) A polynomial cannot have more than one zero.

False.

Justification:

A polynomial can have multiple zeros. The number of zeros a polynomial can have is at most equal to its degree.

Example: Consider the polynomial $p(x) = x^2 - 4$. This is a polynomial of degree 2. The zeros are found by setting $p(x) = 0$, so $x^2 - 4 = 0$, which means $x^2 = 4$. The solutions are $x = 2$ and $x = -2$. This polynomial has two distinct zeros (2 and -2).

Example: Consider the polynomial $p(x) = x(x-1)(x+2) = x^3 + x^2 - 2x$. This is a polynomial of degree 3. The zeros are found by setting $p(x) = 0$, so $x(x-1)(x+2) = 0$. The solutions are $x=0$, $x=1$, and $x=-2$. This polynomial has three distinct zeros (0, 1, and -2).

(vi) The degree of the sum of two polynomials each of degree 5 is always 5.

False.

Justification:

When adding two polynomials, the degree of the sum is typically the highest degree of the individual polynomials. However, if the leading terms (the terms with the highest degree) cancel out, the degree of the sum can be less than the degree of the individual polynomials.

Example 1: Let $p(x) = x^5 + x^2 + 1$ (degree 5) and $q(x) = 2x^5 + x + 3$ (degree 5).

$p(x) + q(x) = (x^5 + x^2 + 1) + (2x^5 + x + 3)$

$= (1+2)x^5 + x^2 + x + (1+3)$

$= 3x^5 + x^2 + x + 4$

The degree of the sum is 5.

Example 2: Let $p(x) = x^5 + x^2 + 1$ (degree 5) and $q(x) = -x^5 + x + 3$ (degree 5).

$p(x) + q(x) = (x^5 + x^2 + 1) + (-x^5 + x + 3)$

$= (1-1)x^5 + x^2 + x + (1+3)$

$= 0x^5 + x^2 + x + 4$

$= x^2 + x + 4$

The degree of the sum is 2, which is less than 5.

(i) Check whether p(x) is a multiple of g(x) or not, where p(x) = x³ – x + 1, g(x) = 2 – 3x.

For $p(x)$ to be a multiple of $g(x)$, $g(x)$ must be a factor of $p(x)$. According to the Factor Theorem, $g(x) = 2 - 3x$ is a factor of $p(x)$ if and only if $p(a) = 0$, where $a$ is the zero of $g(x)$.

To find the zero of $g(x)$, set $g(x) = 0$:

So, the zero of $g(x)$ is $\frac{2}{3}$.

Now, evaluate $p\left(\frac{2}{3}\right)$:

$p(x) = x^3 - x + 1$

$p\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^3 - \left(\frac{2}{3}\right) + 1$

$p\left(\frac{2}{3}\right) = \frac{2^3}{3^3} - \frac{2}{3} + 1$

$p\left(\frac{2}{3}\right) = \frac{8}{27} - \frac{2}{3} + 1$

Find a common denominator, which is 27:

$= \frac{8}{27} - \frac{2 \times 9}{3 \times 9} + \frac{1 \times 27}{1 \times 27}$

$= \frac{8}{27} - \frac{18}{27} + \frac{27}{27}$

$= \frac{8 - 18 + 27}{27}$

$= \frac{-10 + 27}{27}$

$= \frac{17}{27}$

Since $p\left(\frac{2}{3}\right) = \frac{17}{27} \neq 0$, $g(x)$ is not a factor of $p(x)$.

Therefore, $p(x)$ is not a multiple of $g(x)$.

(ii) Check whether g(x) is a factor of p(x) or not, where p(x) = 8x³ – 6x² – 4x + 3, g(x) = $\frac{x}{3}$ - $\frac{1}{4}$.

According to the Factor Theorem, $g(x)$ is a factor of $p(x)$ if and only if $p(a) = 0$, where $a$ is the zero of $g(x)$.

To find the zero of $g(x)$, set $g(x) = 0$:

Multiply both sides by 3:

So, the zero of $g(x)$ is $\frac{3}{4}$.

Now, evaluate $p\left(\frac{3}{4}\right)$:

$p(x) = 8x^3 - 6x^2 - 4x + 3$

$p\left(\frac{3}{4}\right) = 8\left(\frac{3}{4}\right)^3 - 6\left(\frac{3}{4}\right)^2 - 4\left(\frac{3}{4}\right) + 3$

$p\left(\frac{3}{4}\right) = 8\left(\frac{3^3}{4^3}\right) - 6\left(\frac{3^2}{4^2}\right) - 4\left(\frac{3}{4}\right) + 3$

$p\left(\frac{3}{4}\right) = 8\left(\frac{27}{64}\right) - 6\left(\frac{9}{16}\right) - \cancel{4}\left(\frac{3}{\cancel{4}}\right) + 3$

Simplify the terms:

$8\left(\frac{27}{64}\right) = \frac{\cancel{8}^1 \times 27}{\cancel{64}_8} = \frac{27}{8}$

$6\left(\frac{9}{16}\right) = \frac{\cancel{6}^3 \times 9}{\cancel{16}_8} = \frac{27}{8}$

So, $p\left(\frac{3}{4}\right) = \frac{27}{8} - \frac{27}{8} - 3 + 3$

$p\left(\frac{3}{4}\right) = 0 - 0 = 0$

Since $p\left(\frac{3}{4}\right) = 0$, $g(x)$ is a factor of $p(x)$.

Given:

The polynomial $p(x) = x^3 – ax^2 + 2x + a – 1$.

The linear expression $x - a$ is a factor of $p(x)$.

To Find:

The value of $a$.

Solution:

According to the Factor Theorem, if $(x - a)$ is a factor of a polynomial $p(x)$, then $p(a) = 0$.

In this problem, the factor is given as $x - a$. This is already in the form $(x - \text{zero})$, where the zero is $a$.

Substitute $x = a$ into the polynomial $p(x) = x^3 – ax^2 + 2x + a – 1$ and set the result equal to 0.

$p(a) = (a)^3 – a(a)^2 + 2(a) + a – 1$

Since $(x - a)$ is a factor, $p(a) = 0$.

Combine like terms:

Now, solve for $a$. Add 1 to both sides of the equation:

Divide both sides by 3:

Thus, the value of $a$ is $\frac{1}{3}$.

(i) Without actually calculating the cubes, find the value of 48³ – 30³ – 18³.

We can rewrite the expression as $48^3 + (-30)^3 + (-18)^3$.

Consider the identity: If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Let $a = 48$, $b = -30$, and $c = -18$.

Check if $a + b + c = 0$:

$a + b + c = 48 + (-30) + (-18)$

$= 48 - 30 - 18$

$= 48 - (30 + 18)$

$= 48 - 48$

$= 0$

Since $a + b + c = 0$, we can apply the identity $a^3 + b^3 + c^3 = 3abc$.

$48^3 + (-30)^3 + (-18)^3 = 3 \times (48) \times (-30) \times (-18)$

$= 3 \times 48 \times (30 \times 18)$

Calculate the products:

$3 \times 48 = 144$

$30 \times 18 = 540$

$= 144 \times 540$

$144 \times 540 = 77760$

So, $48^3 – 30^3 – 18^3 = 77760$.

(ii) Without finding the cubes, factorise (x – y)³ + (y – z)³ + (z – x)³.

Consider the terms inside the cubes: $(x - y)$, $(y - z)$, and $(z - x)$.

Let $a = x - y$, $b = y - z$, and $c = z - x$.

Check the sum of these terms:

$a + b + c = (x - y) + (y - z) + (z - x)$

$= x - y + y - z + z - x$

$= (x - x) + (-y + y) + (-z + z)$

$= 0 + 0 + 0$

$= 0$

Since $a + b + c = 0$, we can use the identity: If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Substitute $a = x - y$, $b = y - z$, and $c = z - x$ back into the identity:

$(x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x)$

Thus, the factorization is $3(x - y)(y - z)(z - x)$.

To classify a polynomial by the number of variables, we identify the different variables present in the expression.

(i) x² + x + 1

The only variable present is $x$.

This is a polynomial in one variable (x).

(ii) y³ – 5y

The only variable present is $y$.

This is a polynomial in one variable (y).

(iii) xy + yz + zx

The variables present are $x$, $y$, and $z$.

This is a polynomial in three variables (x, y, and z).

(iv) x² – 2xy + y² + 1

The variables present are $x$ and $y$.

This is a polynomial in two variables (x and y).

The degree of a polynomial is the highest exponent of the variable with a non-zero coefficient.

(i) 2x – 1

The polynomial is $2x^1 - 1x^0$.

The powers of the variable $x$ are 1 and 0. The term with the highest power is $2x^1$.

The highest exponent is 1.

The degree of the polynomial is 1.

(ii) –10

The polynomial is $-10$. This is a constant polynomial.

A constant polynomial can be written as $-10x^0$.

The highest exponent of the variable with a non-zero coefficient is 0.

The degree of the polynomial is 0.

(iii) x³ – 9x + 3x⁵

The polynomial is $x^3 - 9x^1 + 3x^5$.

The powers of the variable $x$ are 3, 1, and 5.

Arrange the terms in descending order of exponents: $3x^5 + x^3 - 9x^1$.

The highest exponent is 5.

The degree of the polynomial is 5.

(iv) y³ (1 – y⁴)

First, expand the expression by distributing $y^3$:

$y^3 (1 – y^4) = y^3 \times 1 - y^3 \times y^4$

Using the exponent rule $a^m \times a^n = a^{m+n}$:

$= y^3 - y^{3+4}$

$= y^3 - y^7$

The polynomial is $y^3 - y^7$. Written in descending order of exponents: $-y^7 + y^3$.

The powers of the variable $y$ are 7 and 3.

The highest exponent is 7.

The degree of the polynomial is 7.

The given polynomial is $\frac{x^3 \;+\; 2x \;+\; 1}{5} - \frac{7}{2}x^2 - x^6$.

We can rewrite this polynomial by separating the terms in the first fraction:

$= \frac{x^3}{5} + \frac{2x}{5} + \frac{1}{5} - \frac{7}{2}x^2 - x^6$

Rewrite each term with its coefficient and variable part explicitly:

$= \frac{1}{5}x^3 + \frac{2}{5}x - \frac{7}{2}x^2 + \frac{1}{5} - 1x^6$

Rearrange the terms in descending order of the powers of $x$:

$= -1x^6 + \frac{1}{5}x^3 - \frac{7}{2}x^2 + \frac{2}{5}x + \frac{1}{5}$

Now, let's answer the questions:

(i) the degree of the polynomial

The degree is the highest exponent of $x$ with a non-zero coefficient. In the rearranged polynomial, the powers of $x$ are 6, 3, 2, 1, and 0 (for the constant term).

The highest exponent is 6.

The degree of the polynomial is 6.

(ii) the coefficient of x³

The term with $x^3$ is $\frac{1}{5}x^3$.

The coefficient of $x^3$ is the numerical factor multiplying $x^3$.

The coefficient of $x^3$ is $\frac{1}{5}$.

(iii) the coefficient of x⁶

The term with $x^6$ is $-1x^6$.

The coefficient of $x^6$ is the numerical factor multiplying $x^6$.

The coefficient of $x^6$ is -1.

(iv) the constant term

The constant term is the term that does not have a variable. In the rearranged polynomial, the term with no variable is $\frac{1}{5}$.

The constant term is $\frac{1}{5}$.

(i) The given expression is $\frac{\pi}{6}x + x^2 - 1$.

We can rewrite this in standard polynomial form as $1 \cdot x^2 + \frac{\pi}{6}x - 1$.

The coefficient of $x^2$ is the number multiplying the $x^2$ term.

In this case, the term is $x^2$, which is the same as $1 \cdot x^2$.

Therefore, the coefficient of $x^2$ is $1$.

(ii) The given expression is $3x - 5$.

We can rewrite this expression including an $x^2$ term with a zero coefficient: $0 \cdot x^2 + 3x - 5$.

The coefficient of $x^2$ is the number multiplying the $x^2$ term.

In this case, the coefficient of $x^2$ is $0$.

(iii) The given expression is $(x - 1) (3x - 4)$.

First, we need to expand this expression:

$(x - 1)(3x - 4) = x(3x - 4) - 1(3x - 4)$

$= 3x^2 - 4x - 3x + 4$

Combine the like terms:

$= 3x^2 + (-4x - 3x) + 4$

$= 3x^2 - 7x + 4$

The term containing $x^2$ is $3x^2$.

The coefficient of $x^2$ is the number multiplying $x^2$.

Therefore, the coefficient of $x^2$ is $3$.

(iv) The given expression is $(2x - 5) (2x^2 - 3x + 1)$.

First, we need to expand this expression:

$(2x - 5)(2x^2 - 3x + 1) = 2x(2x^2 - 3x + 1) - 5(2x^2 - 3x + 1)$

$= (2x \cdot 2x^2) + (2x \cdot -3x) + (2x \cdot 1) + (-5 \cdot 2x^2) + (-5 \cdot -3x) $$ + (-5 \cdot 1)$

$= 4x^3 - 6x^2 + 2x - 10x^2 + 15x - 5$

Combine the like terms:

$= 4x^3 + (-6x^2 - 10x^2) + (2x + 15x) - 5$

$= 4x^3 - 16x^2 + 17x - 5$

The term containing $x^2$ is $-16x^2$.

The coefficient of $x^2$ is the number multiplying $x^2$.

Therefore, the coefficient of $x^2$ is $-16$.

The classification of polynomials is based on their degree, which is the highest power of the variable in the polynomial.

Degree 1: Linear polynomial

Degree 2: Quadratic polynomial

Degree 3: Cubic polynomial

Degree 0 (non-zero constant): Constant polynomial

(i) $2 – x^2 + x^3$

The highest power of the variable $x$ is $3$.

Therefore, it is a cubic polynomial.

(ii) $3x^3$

The highest power of the variable $x$ is $3$.

Therefore, it is a cubic polynomial.

(iii) $5t – \sqrt{7}$

The highest power of the variable $t$ is $1$.

Therefore, it is a linear polynomial.

(iv) $4 – 5y^2$

The highest power of the variable $y$ is $2$.

Therefore, it is a quadratic polynomial.

(v) $3$

This is a constant term, which can be written as $3x^0$.

The highest power of the variable is $0$.

Therefore, it is a constant polynomial.

(vi) $2 + x$

The highest power of the variable $x$ is $1$.

Therefore, it is a linear polynomial.

(vii) $y^3 – y$

The highest power of the variable $y$ is $3$.

Therefore, it is a cubic polynomial.

(viii) $1 + x + x^2$

The highest power of the variable $x$ is $2$.

Therefore, it is a quadratic polynomial.

(ix) $t^2$

The highest power of the variable $t$ is $2$.

Therefore, it is a quadratic polynomial.

(x) $\sqrt{2}x – 1$

The highest power of the variable $x$ is $1$.

Therefore, it is a linear polynomial.

(i) A monomial is a polynomial with only one term. The degree of a polynomial is the highest power of the variable.

We need a monomial of degree 1. This means it should have one term and the highest power of the variable should be 1.

An example is a term like $ax$, where $a$ is a non-zero constant.

Example: $5x$ (Here, the term is $5x$ and the degree is 1).

(ii) A binomial is a polynomial with exactly two terms. The degree of a polynomial is the highest power of the variable.

We need a binomial of degree 20. This means it should have two terms and the highest power of the variable in either term should be 20.

Example: One term with a variable raised to the power of 20 and another term with a different power (less than 20) or a constant.

Example: $x^{20} + 9$ (Here, there are two terms $x^{20}$ and $9$, and the highest power is 20).

(iii) A trinomial is a polynomial with exactly three terms. The degree of a polynomial is the highest power of the variable.

We need a trinomial of degree 2. This means it should have three terms and the highest power of the variable in any term should be 2.

Example: One term with a variable raised to the power of 2, another term with a variable raised to the power of 1, and a third term as a constant.

Example: $2x^2 + 3x - 7$ (Here, there are three terms $2x^2$, $3x$, and $-7$, and the highest power is 2).

Let the given polynomial be $P(x) = 3x^3 - 4x^2 + 7x - 5$.

Case 1: When $x = 3$

Substitute $x = 3$ into the polynomial:

$P(3) = 3(3)^3 - 4(3)^2 + 7(3) - 5$

$P(3) = 3(27) - 4(9) + 21 - 5$

$P(3) = 81 - 36 + 21 - 5$

$P(3) = (81 + 21) - (36 + 5)$

$P(3) = 102 - 41$

$P(3) = 61$

So, the value of the polynomial when $x = 3$ is $61$.

Case 2: When $x = -3$

Substitute $x = -3$ into the polynomial:

$P(-3) = 3(-3)^3 - 4(-3)^2 + 7(-3) - 5$

Recall that $(-3)^3 = -27$ and $(-3)^2 = 9$.

$P(-3) = 3(-27) - 4(9) + (-21) - 5$

$P(-3) = -81 - 36 - 21 - 5$

$P(-3) = -(81 + 36 + 21 + 5)$

$P(-3) = -(117 + 21 + 5)$

$P(-3) = -(138 + 5)$

$P(-3) = -143$

So, the value of the polynomial when $x = -3$ is $-143$.

The given polynomial is $p(x) = x^2 - 4x + 3$.

We need to evaluate $p(2) - p(-1) + p(\frac{1}{2})$.

First, let's find the value of $p(2)$:

Substitute $x = 2$ into the polynomial:

$p(2) = (2)^2 - 4(2) + 3$

$p(2) = 4 - 8 + 3$

$p(2) = -4 + 3$

$p(2) = -1$

Next, let's find the value of $p(-1)$:

Substitute $x = -1$ into the polynomial:

$p(-1) = (-1)^2 - 4(-1) + 3$

$p(-1) = 1 - (-4) + 3$

$p(-1) = 1 + 4 + 3$

$p(-1) = 5 + 3$

$p(-1) = 8$

Now, let's find the value of $p(\frac{1}{2})$:

Substitute $x = \frac{1}{2}$ into the polynomial:

$p(\frac{1}{2}) = (\frac{1}{2})^2 - 4(\frac{1}{2}) + 3$

$p(\frac{1}{2}) = \frac{1}{4} - \frac{4}{2} + 3$

$p(\frac{1}{2}) = \frac{1}{4} - 2 + 3$

$p(\frac{1}{2}) = \frac{1}{4} + 1$

To add $\frac{1}{4}$ and $1$, we find a common denominator:

$p(\frac{1}{2}) = \frac{1}{4} + \frac{4}{4}$

$p(\frac{1}{2}) = \frac{1+4}{4}$

$p(\frac{1}{2}) = \frac{5}{4}$

Finally, we evaluate $p(2) - p(-1) + p(\frac{1}{2})$:

$p(2) - p(-1) + p(\frac{1}{2}) = -1 - 8 + \frac{5}{4}$

$= -9 + \frac{5}{4}$

To combine $-9$ and $\frac{5}{4}$, we find a common denominator:

$= -\frac{9 \cdot 4}{4} + \frac{5}{4}$

$= -\frac{36}{4} + \frac{5}{4}$

$= \frac{-36 + 5}{4}$

$= \frac{-31}{4}$

The value of $p(2) - p(-1) + p(\frac{1}{2})$ is $-\frac{31}{4}$.

(i) Given polynomial is $p(x) = 10x - 4x^2 - 3$.

We need to find $p(0)$, $p(1)$, and $p(-2)$.

Find $p(0)$:

Substitute $x = 0$ into the polynomial:

$p(0) = 10(0) - 4(0)^2 - 3$

$p(0) = 0 - 4(0) - 3$

$p(0) = 0 - 0 - 3$

$p(0) = -3$

Find $p(1)$:

Substitute $x = 1$ into the polynomial:

$p(1) = 10(1) - 4(1)^2 - 3$

$p(1) = 10 - 4(1) - 3$

$p(1) = 10 - 4 - 3$

$p(1) = 6 - 3$

$p(1) = 3$

Find $p(-2)$:

Substitute $x = -2$ into the polynomial:

$p(-2) = 10(-2) - 4(-2)^2 - 3$

Recall that $(-2)^2 = 4$.

$p(-2) = -20 - 4(4) - 3$

$p(-2) = -20 - 16 - 3$

$p(-2) = -36 - 3$

$p(-2) = -39$

(ii) Given polynomial is $p(y) = (y + 2)(y - 2)$.

We can expand this using the identity $(a+b)(a-b) = a^2 - b^2$:

$p(y) = y^2 - 2^2$

$p(y) = y^2 - 4$

We need to find $p(0)$, $p(1)$, and $p(-2)$.

Find $p(0)$:

Substitute $y = 0$ into the polynomial:

$p(0) = (0)^2 - 4$

$p(0) = 0 - 4$

$p(0) = -4$

Find $p(1)$:

Substitute $y = 1$ into the polynomial:

$p(1) = (1)^2 - 4$

$p(1) = 1 - 4$

$p(1) = -3$

Find $p(-2)$:

Substitute $y = -2$ into the polynomial:

$p(-2) = (-2)^2 - 4$

$p(-2) = 4 - 4$

$p(-2) = 0$

A number 'a' is a zero of a polynomial $p(x)$ if $p(a) = 0$. We need to evaluate the polynomial at the given value to verify if it is a zero.

(i) Given polynomial is $p(x) = x - 3$. We need to check if $-3$ is a zero.

Substitute $x = -3$:

$p(-3) = (-3) - 3$

$p(-3) = -6$

Since $p(-3) \neq 0$, $-3$ is not a zero of $x - 3$.

Therefore, the statement is False.

(ii) Given polynomial is $p(x) = 3x + 1$. We need to check if $-\frac{1}{3}$ is a zero.

Substitute $x = -\frac{1}{3}$:

$p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1$

$p(-\frac{1}{3}) = -1 + 1$

$p(-\frac{1}{3}) = 0$

Since $p(-\frac{1}{3}) = 0$, $-\frac{1}{3}$ is a zero of $3x + 1$.

Therefore, the statement is True.

(iii) Given polynomial is $p(y) = 4 - 5y$. We need to check if $\frac{-4}{5}$ is a zero.

Substitute $y = \frac{-4}{5}$:

$p(\frac{-4}{5}) = 4 - 5(\frac{-4}{5})$

$p(\frac{-4}{5}) = 4 - (\cancel{5} \cdot \frac{-4}{\cancel{5}})$

$p(\frac{-4}{5}) = 4 - (-4)$

$p(\frac{-4}{5}) = 4 + 4$

$p(\frac{-4}{5}) = 8$

Since $p(\frac{-4}{5}) \neq 0$, $\frac{-4}{5}$ is not a zero of $4 - 5y$.

Therefore, the statement is False.

(iv) Given polynomial is $p(t) = t^2 - 2t$. We need to check if $0$ and $2$ are zeroes.

First, check for $t = 0$:

$p(0) = (0)^2 - 2(0)$

$p(0) = 0 - 0$

$p(0) = 0$

Since $p(0) = 0$, $0$ is a zero of $t^2 - 2t$.

Next, check for $t = 2$:

$p(2) = (2)^2 - 2(2)$

$p(2) = 4 - 4$

$p(2) = 0$

Since $p(2) = 0$, $2$ is a zero of $t^2 - 2t$.

Since both $p(0) = 0$ and $p(2) = 0$, both $0$ and $2$ are zeroes of the polynomial.

Therefore, the statement is True.

(v) Given polynomial is $p(y) = y^2 + y - 6$. We need to check if $-3$ is a zero.

Substitute $y = -3$:

$p(-3) = (-3)^2 + (-3) - 6$

$p(-3) = 9 - 3 - 6$

$p(-3) = 6 - 6$

$p(-3) = 0$

Since $p(-3) = 0$, $-3$ is a zero of $y^2 + y - 6$.

Therefore, the statement is True.

To find the zero of a polynomial, we set the polynomial equal to zero and solve for the variable.

(i) Given polynomial is $p(x) = x - 4$.

To find the zero, set $p(x) = 0$:

Add 4 to both sides:

The zero of the polynomial $p(x) = x - 4$ is $4$.

(ii) Given polynomial is $g(x) = 3 - 6x$.

To find the zero, set $g(x) = 0$:

Add $6x$ to both sides:

Divide both sides by 6:

Simplify the fraction:

The zero of the polynomial $g(x) = 3 - 6x$ is $\frac{1}{2}$.

(iii) Given polynomial is $q(x) = 2x - 7$.

To find the zero, set $q(x) = 0$:

Add 7 to both sides:

Divide both sides by 2:

The zero of the polynomial $q(x) = 2x - 7$ is $\frac{7}{2}$.

(iv) Given polynomial is $h(y) = 2y$.

To find the zero, set $h(y) = 0$:

Divide both sides by 2:

The zero of the polynomial $h(y) = 2y$ is $0$.

The given polynomial is $p(x) = (x - 2)^2 - (x + 2)^2$.

To find the zeroes of the polynomial, we set $p(x) = 0$.

$(x - 2)^2 - (x + 2)^2 = 0$

We can expand the terms or use the difference of squares identity, $a^2 - b^2 = (a - b)(a + b)$.

Let $a = (x - 2)$ and $b = (x + 2)$.

So, $p(x) = [(x - 2) - (x + 2)][(x - 2) + (x + 2)]$

Simplify the terms inside the brackets:

$(x - 2 - x - 2)(x - 2 + x + 2) = 0$

Combine like terms in each bracket:

$(x - x - 2 - 2)(x + x - 2 + 2) = 0$

$(-4)(2x) = 0$

Multiply the terms:

$-8x = 0$

Now, solve for $x$:

Alternatively, we could expand the squares directly:

$(x - 2)^2 = x^2 - 2(x)(2) + (-2)^2 = x^2 - 4x + 4$

$(x + 2)^2 = x^2 + 2(x)(2) + (2)^2 = x^2 + 4x + 4$

So, $p(x) = (x^2 - 4x + 4) - (x^2 + 4x + 4)$

$p(x) = x^2 - 4x + 4 - x^2 - 4x - 4$

Combine like terms:

$p(x) = (x^2 - x^2) + (-4x - 4x) + (4 - 4)$

$p(x) = 0x^2 - 8x + 0$

$p(x) = -8x$

Setting $p(x) = 0$:

The zero of the polynomial $p(x) = (x - 2)^2 - (x + 2)^2$ is $0$.

Given:

The first polynomial (Dividend) is $p(x) = x^4 + 1$

The second polynomial (Divisor) is $g(x) = x - 1$

To Find:

The quotient and the remainder using the actual division method.

Solution:

To perform the actual division, we write the dividend $x^4 + 1$ in standard form by including the missing terms with zero coefficients: $x^4 + 0x^3 + 0x^2 + 0x + 1$.

The division process is shown below:

$\begin{array}{r} x^3 + x^2 + x + 1 \phantom{x^4+0x^3)} \\ x-1{\overline{\smash{\big)}\,x^4 + 0x^3 + 0x^2 + 0x + 1 \phantom{)}}} \\ \underline{-~\phantom{(}(x^4 - x^3)\phantom{+0x^2+0x+1)}} \\ x^3 + 0x^2 \phantom{+0x+1)} \\ \underline{-~\phantom{(}(x^3 - x^2)\phantom{+0x+1)}} \\ x^2 + 0x \phantom{+1)} \\ \underline{-~\phantom{(}(x^2 - x)\phantom{+1)}} \\ x + 1 \phantom{)} \\ \underline{-~\phantom{(}(x - 1)} \\ 2 \phantom{)} \end{array}$

From the above long division, we observe that:

The Quotient is $x^3 + x^2 + x + 1$

The Remainder is $2$

Alternate Method (Using Remainder Theorem):

According to the Remainder Theorem, if a polynomial $p(x)$ is divided by $(x - a)$, then the remainder is $p(a)$.

Here, the divisor is $x - 1$, so we substitute $x = 1$ in the dividend $p(x) = x^4 + 1$:

$p(1) = (1)^4 + 1$

$p(1) = 1 + 1$

$p(1) = 2$

The remainder obtained is $2$, which matches the result of our actual division.

According to the Remainder Theorem, if a polynomial $p(x)$ is divided by a linear polynomial $g(x) = x - a$, then the remainder is $p(a)$. If $g(x) = bx - a$, the zero is $x = \frac{a}{b}$, and the remainder is $p(\frac{a}{b})$.

(i) Given $p(x) = x^3 – 2x^2 – 4x – 1$ and $g(x) = x + 1$.

To find the zero of $g(x)$, set $g(x) = 0$:

According to the Remainder Theorem, the remainder when $p(x)$ is divided by $x + 1$ is $p(-1)$.

Substitute $x = -1$ into $p(x)$:

$p(-1) = (-1)^3 - 2(-1)^2 - 4(-1) - 1$

$p(-1) = -1 - 2(1) - (-4) - 1$

$p(-1) = -1 - 2 + 4 - 1$

$p(-1) = -3 + 4 - 1$

$p(-1) = 1 - 1$

$p(-1) = 0$

The remainder is $0$.

(ii) Given $p(x) = x^3 – 3x^2 + 4x + 50$ and $g(x) = x - 3$.

To find the zero of $g(x)$, set $g(x) = 0$:

According to the Remainder Theorem, the remainder when $p(x)$ is divided by $x - 3$ is $p(3)$.

Substitute $x = 3$ into $p(x)$:

$p(3) = (3)^3 - 3(3)^2 + 4(3) + 50$

$p(3) = 27 - 3(9) + 12 + 50$

$p(3) = 27 - 27 + 12 + 50$

$p(3) = 0 + 12 + 50$

$p(3) = 62$

The remainder is $62$.

(iii) Given $p(x) = 4x^3 – 12x^2 + 14x – 3$ and $g(x) = 2x - 1$.

To find the zero of $g(x)$, set $g(x) = 0$:

According to the Remainder Theorem, the remainder when $p(x)$ is divided by $2x - 1$ is $p(\frac{1}{2})$.

Substitute $x = \frac{1}{2}$ into $p(x)$:

$p(\frac{1}{2}) = 4(\frac{1}{2})^3 - 12(\frac{1}{2})^2 + 14(\frac{1}{2}) - 3$

$p(\frac{1}{2}) = 4(\frac{1}{8}) - 12(\frac{1}{4}) + \cancel{14}^7(\frac{1}{\cancel{2}_1}) - 3$

$p(\frac{1}{2}) = \frac{\cancel{4}^1}{\cancel{8}_2} - \frac{\cancel{12}^3}{\cancel{4}_1} + 7 - 3$

$p(\frac{1}{2}) = \frac{1}{2} - 3 + 7 - 3$

$p(\frac{1}{2}) = \frac{1}{2} + 4 - 3$

$p(\frac{1}{2}) = \frac{1}{2} + 1$

$p(\frac{1}{2}) = \frac{1}{2} + \frac{2}{2}$

$p(\frac{1}{2}) = \frac{1+2}{2}$

$p(\frac{1}{2}) = \frac{3}{2}$

The remainder is $\frac{3}{2}$.

(iv) Given $p(x) = x^3 – 6x^2 + 2x – 4$ and $g(x) = 1 – \frac{3}{2}x$.

To find the zero of $g(x)$, set $g(x) = 0$:

Multiply both sides by $\frac{2}{3}$:

According to the Remainder Theorem, the remainder when $p(x)$ is divided by $1 – \frac{3}{2}x$ is $p(\frac{2}{3})$.

Substitute $x = \frac{2}{3}$ into $p(x)$:

$p(\frac{2}{3}) = (\frac{2}{3})^3 - 6(\frac{2}{3})^2 + 2(\frac{2}{3}) - 4$

$p(\frac{2}{3}) = \frac{2^3}{3^3} - 6(\frac{2^2}{3^2}) + \frac{4}{3} - 4$

$p(\frac{2}{3}) = \frac{8}{27} - 6(\frac{4}{9}) + \frac{4}{3} - 4$

$p(\frac{2}{3}) = \frac{8}{27} - \frac{\cancel{6}^2 \cdot 4}{\cancel{9}_3} + \frac{4}{3} - 4$

$p(\frac{2}{3}) = \frac{8}{27} - \frac{8}{3} + \frac{4}{3} - 4$

Combine the terms with denominator 3:

$p(\frac{2}{3}) = \frac{8}{27} + \frac{4 - 8}{3} - 4$

$p(\frac{2}{3}) = \frac{8}{27} + \frac{-4}{3} - 4$

Find a common denominator for the fractions, which is 27.

$p(\frac{2}{3}) = \frac{8}{27} - \frac{4 \cdot 9}{3 \cdot 9} - \frac{4 \cdot 27}{1 \cdot 27}$

$p(\frac{2}{3}) = \frac{8}{27} - \frac{36}{27} - \frac{108}{27}$

$p(\frac{2}{3}) = \frac{8 - 36 - 108}{27}$

$p(\frac{2}{3}) = \frac{-28 - 108}{27}$

$p(\frac{2}{3}) = \frac{-136}{27}$

The remainder is $-\frac{136}{27}$.

A polynomial $p(x)$ is a multiple of a polynomial $g(x)$ if and only if the remainder when $p(x)$ is divided by $g(x)$ is zero. We can use the Remainder Theorem to find the remainder.

(i) Given $p(x) = x^3 – 5x^2 + 4x – 3$ and $g(x) = x – 2$.

To check if $p(x)$ is a multiple of $g(x)$, we find the zero of $g(x)$ and evaluate $p(x)$ at that value.

Set $g(x) = 0$:

Now, evaluate $p(2)$:

$p(2) = (2)^3 - 5(2)^2 + 4(2) - 3$

$p(2) = 8 - 5(4) + 8 - 3$

$p(2) = 8 - 20 + 8 - 3$

$p(2) = 16 - 20 - 3$

$p(2) = -4 - 3$

$p(2) = -7$

Since the remainder $p(2) = -7 \neq 0$, $p(x)$ is not a multiple of $g(x)$.

Therefore, the statement is False.

(ii) Given $p(x) = 2x^3 – 11x^2 – 4x + 5$ and $g(x) = 2x + 1$.

To check if $p(x)$ is a multiple of $g(x)$, we find the zero of $g(x)$ and evaluate $p(x)$ at that value.

Set $g(x) = 0$:

Now, evaluate $p(-\frac{1}{2})$:

$p(-\frac{1}{2}) = 2(-\frac{1}{2})^3 - 11(-\frac{1}{2})^2 - 4(-\frac{1}{2}) + 5$

$p(-\frac{1}{2}) = 2(-\frac{1}{8}) - 11(\frac{1}{4}) - (\cancel{4}^2)(-\frac{1}{\cancel{2}_1}) + 5$

$p(-\frac{1}{2}) = -\frac{\cancel{2}^1}{\cancel{8}_4} - \frac{11}{4} - (-2) + 5$

$p(-\frac{1}{2}) = -\frac{1}{4} - \frac{11}{4} + 2 + 5$

$p(-\frac{1}{2}) = \frac{-1 - 11}{4} + 7$

$p(-\frac{1}{2}) = \frac{-12}{4} + 7$

$p(-\frac{1}{2}) = -3 + 7$

$p(-\frac{1}{2}) = 4$

Since the remainder $p(-\frac{1}{2}) = 4 \neq 0$, $p(x)$ is not a multiple of $g(x)$.

Therefore, the statement is False.

According to the Factor Theorem, a polynomial $g(x)$ is a factor of a polynomial $p(x)$ if and only if $p(a) = 0$, where $a$ is the zero of $g(x)$ (i.e., $g(a) = 0$).

(i) Given $p(x) = 69 + 11x – x^2 + x^3$ and $g(x) = x + 3$.

First, find the zero of $g(x)$. Set $g(x) = 0$:

Now, evaluate $p(x)$ at $x = -3$. If $p(-3) = 0$, then $x + 3$ is a factor of $p(x)$.

Substitute $x = -3$ into $p(x)$:

$p(-3) = 69 + 11(-3) - (-3)^2 + (-3)^3$

$p(-3) = 69 - 33 - (9) + (-27)$

$p(-3) = 69 - 33 - 9 - 27$

$p(-3) = (69 - 33) - 9 - 27$

$p(-3) = 36 - 9 - 27$

$p(-3) = 27 - 27$

$p(-3) = 0$

Since $p(-3) = 0$, by the Factor Theorem, $x + 3$ is a factor of $p(x)$.

(ii) Given $p(x) = x + 2x^3 – 9x^2 + 12$ and $g(x) = 2x – 3$.

First, find the zero of $g(x)$. Set $g(x) = 0$:

Now, evaluate $p(x)$ at $x = \frac{3}{2}$. If $p(\frac{3}{2}) = 0$, then $2x - 3$ is a factor of $p(x)$.

Substitute $x = \frac{3}{2}$ into $p(x)$:

$p(\frac{3}{2}) = (\frac{3}{2}) + 2(\frac{3}{2})^3 - 9(\frac{3}{2})^2 + 12$

$p(\frac{3}{2}) = \frac{3}{2} + 2(\frac{27}{8}) - 9(\frac{9}{4}) + 12$

$p(\frac{3}{2}) = \frac{3}{2} + \frac{\cancel{2}^1 \cdot 27}{\cancel{8}_4} - \frac{81}{4} + 12$

$p(\frac{3}{2}) = \frac{3}{2} + \frac{27}{4} - \frac{81}{4} + 12$

Find a common denominator for the fractions, which is 4.

$p(\frac{3}{2}) = \frac{3 \cdot 2}{2 \cdot 2} + \frac{27}{4} - \frac{81}{4} + \frac{12 \cdot 4}{1 \cdot 4}$

$p(\frac{3}{2}) = \frac{6}{4} + \frac{27}{4} - \frac{81}{4} + \frac{48}{4}$

$p(\frac{3}{2}) = \frac{6 + 27 - 81 + 48}{4}$

$p(\frac{3}{2}) = \frac{33 - 81 + 48}{4}$

$p(\frac{3}{2}) = \frac{-48 + 48}{4}$

$p(\frac{3}{2}) = \frac{0}{4}$

$p(\frac{3}{2}) = 0$

Since $p(\frac{3}{2}) = 0$, by the Factor Theorem, $2x - 3$ is a factor of $p(x)$.

According to the Factor Theorem, a linear polynomial $(x - a)$ is a factor of a polynomial $p(x)$ if and only if $p(a) = 0$.

In this case, we are checking if $(x - 2)$ is a factor. The zero of $(x - 2)$ is found by setting $x - 2 = 0$, which gives $x = 2$.

So, we need to evaluate each given polynomial at $x = 2$. If the value of the polynomial at $x = 2$ is $0$, then $(x - 2)$ is a factor.

(i) Let $p(x) = 3x^2 + 6x - 24$.

Evaluate $p(2)$:

$p(2) = 3(2)^2 + 6(2) - 24$

$p(2) = 3(4) + 12 - 24$

$p(2) = 12 + 12 - 24$

$p(2) = 24 - 24$

$p(2) = 0$

Since $p(2) = 0$, $(x - 2)$ is a factor of the polynomial $3x^2 + 6x - 24$.

(ii) Let $q(x) = 4x^2 + x - 2$.

Evaluate $q(2)$:

$q(2) = 4(2)^2 + (2) - 2$

$q(2) = 4(4) + 2 - 2$

$q(2) = 16 + 2 - 2$

$q(2) = 16 + 0$

$q(2) = 16$

Since $q(2) = 16 \neq 0$, $(x - 2)$ is not a factor of the polynomial $4x^2 + x - 2$.

Therefore, the polynomial that has $(x - 2)$ as a factor is $3x^2 + 6x - 24$.

According to the Factor Theorem, a linear polynomial $(x - a)$ is a factor of a polynomial $P(x)$ if and only if $P(a) = 0$. In this question, the variable is $p$ and the potential factor is $(p - 1)$. So, $a = 1$.

To show that $p - 1$ is a factor of a polynomial, we need to evaluate the polynomial at $p = 1$. If the result is $0$, then $p - 1$ is a factor.

Part 1: Show that $p - 1$ is a factor of $p^{10} - 1$.

Let $P_1(p) = p^{10} - 1$.

The zero of the linear polynomial $p - 1$ is found by setting $p - 1 = 0$, which gives $p = 1$.

Now, evaluate $P_1(p)$ at $p = 1$:

$P_1(1) = (1)^{10} - 1$

$P_1(1) = 1 - 1$

$P_1(1) = 0$

Since $P_1(1) = 0$, by the Factor Theorem, $(p - 1)$ is a factor of $p^{10} - 1$.

Part 2: Show that $p - 1$ is a factor of $p^{11} - 1$.

Let $P_2(p) = p^{11} - 1$.

The zero of the linear polynomial $p - 1$ is $p = 1$.

Now, evaluate $P_2(p)$ at $p = 1$:

$P_2(1) = (1)^{11} - 1$

$P_2(1) = 1 - 1$

$P_2(1) = 0$

Since $P_2(1) = 0$, by the Factor Theorem, $(p - 1)$ is a factor of $p^{11} - 1$.

Therefore, $p - 1$ is a factor of both $p^{10} - 1$ and $p^{11} - 1$.

Let the given polynomial be $p(x) = x^3 - 2mx^2 + 16$.

The polynomial $p(x)$ is divisible by $x + 2$ if and only if $(x + 2)$ is a factor of $p(x)$.

According to the Factor Theorem, $(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$.

In this case, the divisor is $x + 2$, which can be written as $x - (-2)$. So, $a = -2$.

For $p(x)$ to be divisible by $x + 2$, the remainder when $p(x)$ is divided by $x + 2$ must be $0$. By the Remainder Theorem, this remainder is $p(-2)$.

Therefore, we must have $p(-2) = 0$.

Substitute $x = -2$ into the polynomial $p(x)$:

$p(-2) = (-2)^3 - 2m(-2)^2 + 16$

Calculate the powers:

$(-2)^3 = -8$

$(-2)^2 = 4$

Substitute these values back:

$p(-2) = -8 - 2m(4) + 16$

$p(-2) = -8 - 8m + 16$

Combine the constant terms:

$p(-2) = (-8 + 16) - 8m$

$p(-2) = 8 - 8m$

Since $p(x)$ is divisible by $x + 2$, the remainder must be $0$. Set $p(-2) = 0$:

Solve for $m$. Add $8m$ to both sides:

Divide both sides by 8:

Thus, for the polynomial $x^3 - 2mx^2 + 16$ to be divisible by $x + 2$, the value of $m$ must be $1$.

Let the given polynomial be $p(x) = x^5 – 4a^2x^3 + 2x + 2a + 3$.

We are given that $x + 2a$ is a factor of $p(x)$.

According to the Factor Theorem, if $(x - c)$ is a factor of a polynomial $p(x)$, then $p(c) = 0$.

In this case, the factor is $x + 2a$, which can be written as $x - (-2a)$. So, $c = -2a$.

For $x + 2a$ to be a factor of $p(x)$, the value of the polynomial at $x = -2a$ must be $0$.

Thus, we must have $p(-2a) = 0$.

Substitute $x = -2a$ into the polynomial $p(x)$:

$p(-2a) = (-2a)^5 - 4a^2(-2a)^3 + 2(-2a) + 2a + 3$

Calculate the terms:

$(-2a)^5 = (-2)^5 \cdot a^5 = -32a^5$

$(-2a)^3 = (-2)^3 \cdot a^3 = -8a^3$

Substitute these values back into the expression for $p(-2a)$:

$p(-2a) = -32a^5 - 4a^2(-8a^3) + 2(-2a) + 2a + 3$

$p(-2a) = -32a^5 + 32a^5 - 4a + 2a + 3$

Combine the like terms:

$p(-2a) = (-32a^5 + 32a^5) + (-4a + 2a) + 3$

$p(-2a) = 0a^5 - 2a + 3$

$p(-2a) = -2a + 3$

Since $x + 2a$ is a factor, $p(-2a)$ must be equal to 0.

Solve for $a$. Subtract 3 from both sides:

Divide both sides by -2:

The value of $a$ for which $x + 2a$ is a factor of the given polynomial is $\frac{3}{2}$.

Let the given polynomial be $p(x) = 8x^4 + 4x^3 – 16x^2 + 10x + m$.

We are given that $2x - 1$ is a factor of $p(x)$.

According to the Factor Theorem, a polynomial $g(x)$ is a factor of $p(x)$ if and only if $p(a) = 0$, where $a$ is the zero of $g(x)$.

First, find the zero of the divisor $g(x) = 2x - 1$. Set $g(x) = 0$:

Since $2x - 1$ is a factor of $p(x)$, by the Factor Theorem, the value of $p(x)$ at $x = \frac{1}{2}$ must be $0$.

Thus, $p(\frac{1}{2}) = 0$.

Substitute $x = \frac{1}{2}$ into the polynomial $p(x)$:

$p(\frac{1}{2}) = 8(\frac{1}{2})^4 + 4(\frac{1}{2})^3 - 16(\frac{1}{2})^2 + 10(\frac{1}{2}) + m$

Calculate the powers of $\frac{1}{2}$:

$(\frac{1}{2})^4 = \frac{1}{16}$

$(\frac{1}{2})^3 = \frac{1}{8}$

$(\frac{1}{2})^2 = \frac{1}{4}$

Substitute these values back into the expression for $p(\frac{1}{2})$:

$p(\frac{1}{2}) = 8(\frac{1}{16}) + 4(\frac{1}{8}) - 16(\frac{1}{4}) + 10(\frac{1}{2}) + m$

Simplify the terms:

$p(\frac{1}{2}) = \frac{8}{16} + \frac{4}{8} - \frac{16}{4} + \frac{10}{2} + m$

$p(\frac{1}{2}) = \frac{1}{2} + \frac{1}{2} - 4 + 5 + m$

Combine the constant terms:

$p(\frac{1}{2}) = (\frac{1}{2} + \frac{1}{2}) + (-4 + 5) + m$

$p(\frac{1}{2}) = 1 + 1 + m$

$p(\frac{1}{2}) = 2 + m$

Since $p(\frac{1}{2})$ must be $0$:

Solve for $m$:

The value of $m$ for which $2x - 1$ is a factor of the polynomial is $-2$.

Let the given polynomial be $p(x) = ax^3 + x^2 – 2x + 4a – 9$.

We are given that $x + 1$ is a factor of $p(x)$.

According to the Factor Theorem, if $(x - c)$ is a factor of a polynomial $p(x)$, then $p(c) = 0$.

In this case, the factor is $x + 1$, which can be written as $x - (-1)$. So, the zero of the factor is $x = -1$.

For $x + 1$ to be a factor of $p(x)$, the value of the polynomial at $x = -1$ must be $0$.

Thus, we must have $p(-1) = 0$.

Substitute $x = -1$ into the polynomial $p(x)$:

$p(-1) = a(-1)^3 + (-1)^2 - 2(-1) + 4a - 9$

Calculate the terms:

$(-1)^3 = -1$

$(-1)^2 = 1$

Substitute these values back into the expression for $p(-1)$:

$p(-1) = a(-1) + 1 - (-2) + 4a - 9$

$p(-1) = -a + 1 + 2 + 4a - 9$

Combine the like terms (terms with $a$ and constant terms):

$p(-1) = (-a + 4a) + (1 + 2 - 9)$

$p(-1) = 3a + (3 - 9)$

$p(-1) = 3a - 6$

Since $x + 1$ is a factor, $p(-1)$ must be equal to 0.

Solve for $a$. Add 6 to both sides:

Divide both sides by 3:

The value of $a$ is $2$.

(i) We need to factorise the quadratic polynomial $x^2 + 9x + 18$.

We look for two numbers whose sum is the coefficient of $x$ (which is 9) and whose product is the constant term (which is 18).

Let the two numbers be $p$ and $q$. We need $p + q = 9$ and $pq = 18$.

The pairs of factors of 18 are $(1, 18), (2, 9), (3, 6), $$ (-1, -18), (-2, -9), (-3, -6)$.

The pair that adds up to 9 is $(3, 6)$.

So we split the middle term $9x$ as $3x + 6x$.

$x^2 + 9x + 18 = x^2 + 3x + 6x + 18$

Now, we factor by grouping the terms:

$= (x^2 + 3x) + (6x + 18)$

Factor out the common factor from each group:

$= x(x + 3) + 6(x + 3)$

Factor out the common binomial factor $(x + 3)$:

$= (x + 3)(x + 6)$

The factorisation is $(x + 3)(x + 6)$.

(ii) We need to factorise the quadratic polynomial $6x^2 + 7x – 3$.

We look for two numbers whose sum is the coefficient of $x$ (which is 7) and whose product is the product of the coefficient of $x^2$ (6) and the constant term (-3), i.e., $6 \times (-3) = -18$.

Let the two numbers be $p$ and $q$. We need $p + q = 7$ and $pq = -18$.

The pairs of factors of -18 are $(1, -18), (-1, 18), (2, -9), $$ (-2, 9), (3, -6), (-3, 6)$.

The pair that adds up to 7 is $(-2, 9)$.

So we split the middle term $7x$ as $-2x + 9x$.

$6x^2 + 7x – 3 = 6x^2 - 2x + 9x - 3$

Now, we factor by grouping the terms:

$= (6x^2 - 2x) + (9x - 3)$

Factor out the common factor from each group:

$= 2x(3x - 1) + 3(3x - 1)$

Factor out the common binomial factor $(3x - 1)$:

$= (3x - 1)(2x + 3)$

The factorisation is $(3x - 1)(2x + 3)$.

(iii) We need to factorise the quadratic polynomial $2x^2 – 7x – 15$.

We look for two numbers whose sum is the coefficient of $x$ (which is -7) and whose product is the product of the coefficient of $x^2$ (2) and the constant term (-15), i.e., $2 \times (-15) = -30$.

Let the two numbers be $p$ and $q$. We need $p + q = -7$ and $pq = -30$.

The pairs of factors of -30 are $(1, -30), (-1, 30), (2, -15), (-2, 15), $$ (3, -10), (-3, 10), (5, -6), (-5, 6)$.

The pair that adds up to -7 is $(3, -10)$.

So we split the middle term $-7x$ as $3x - 10x$.

$2x^2 – 7x – 15 = 2x^2 + 3x - 10x - 15$

Now, we factor by grouping the terms:

$= (2x^2 + 3x) + (-10x - 15)$

Factor out the common factor from each group. Note that in the second group, we factor out -5.

$= x(2x + 3) - 5(2x + 3)$

Factor out the common binomial factor $(2x + 3)$:

$= (2x + 3)(x - 5)$

The factorisation is $(2x + 3)(x - 5)$.

(iv) We need to factorise the polynomial $84 – 2r – 2r^2$.

First, let's rearrange the terms in descending powers of $r$ and factor out the common numerical factor, which is 2.

$84 – 2r – 2r^2 = -2r^2 - 2r + 84$

Factor out -2:

$= -2(r^2 + r - 42)$

Now we need to factorise the quadratic expression inside the bracket, $r^2 + r - 42$.

We look for two numbers whose sum is the coefficient of $r$ (which is 1) and whose product is the constant term (which is -42).

Let the two numbers be $p$ and $q$. We need $p + q = 1$ and $pq = -42$.

The pairs of factors of -42 are $(1, -42), (-1, 42), (2, -21), (-2, 21), $$ (3, -14), (-3, 14), (6, -7), (-6, 7)$.

The pair that adds up to 1 is $(-6, 7)$.

So we split the middle term $r$ as $-6r + 7r$.

$r^2 + r - 42 = r^2 - 6r + 7r - 42$

Now, we factor by grouping the terms:

$= (r^2 - 6r) + (7r - 42)$

Factor out the common factor from each group:

$= r(r - 6) + 7(r - 6)$

Factor out the common binomial factor $(r - 6)$:

$= (r - 6)(r + 7)$

Substitute this back into the original expression:

$84 – 2r – 2r^2 = -2(r - 6)(r + 7)$

The factorisation is $-2(r - 6)(r + 7)$.

To Factorise:

(i) $2x^3 - 3x^2 - 17x + 30$

(ii) $x^3 - 6x^2 + 11x - 6$

(iii) $x^3 + x^2 - 4x - 4$

(iv) $3x^3 - x^2 - 3x + 1$

Solution (i): $2x^3 - 3x^2 - 17x + 30$

Let $p(x) = 2x^3 - 3x^2 - 17x + 30$.

The constant term is $30$. Factors of $30$ are $\pm 1, \pm 2, \pm 3, \pm 5, \dots$

By trial method, let us check for $x = 2$:

$p(2) = 2(2)^3 - 3(2)^2 - 17(2) + 30$

$p(2) = 2(8) - 3(4) - 34 + 30$

$p(2) = 16 - 12 - 34 + 30$

Since $p(2) = 0$, by the Factor Theorem, $(x - 2)$ is a factor of $p(x)$.

Now, we divide $p(x)$ by $(x - 2)$ to find the other factors:

$\begin{array}{r} 2x^2 + x - 15 \phantom{2x^3-3x^2)} \\ x-2{\overline{\smash{\big)}\,2x^3-3x^2-17x+30\phantom{)}}} \\ \underline{-~\phantom{(}(2x^3-4x^2)\phantom{-17x+30)}} \\ x^2-17x \phantom{+30)} \\ \underline{-~\phantom{(}(x^2-2x)\phantom{+30)}} \\ -15x+30 \phantom{)} \\ \underline{-~\phantom{(}(-15x+30)} \\ 0 \phantom{)} \end{array}$

Now, we factorise the quadratic quotient $2x^2 + x - 15$ by splitting the middle term:

$2x^2 + 6x - 5x - 15$

$2x(x + 3) - 5(x + 3)$

$(2x - 5)(x + 3)$

Therefore, $2x^3 - 3x^2 - 17x + 30 = (x - 2)(x + 3)(2x - 5)$.

Solution (ii): $x^3 - 6x^2 + 11x - 6$

Let $p(x) = x^3 - 6x^2 + 11x - 6$.

The constant term is $-6$. Factors of $-6$ are $\pm 1, \pm 2, \pm 3, \pm 6$.

By trial method, let us check for $x = 1$:

$p(1) = (1)^3 - 6(1)^2 + 11(1) - 6$

Since $p(1) = 0$, by the Factor Theorem, $(x - 1)$ is a factor of $p(x)$.

Now, we divide $p(x)$ by $(x - 1)$:

$\begin{array}{r} x^2 - 5x + 6 \phantom{x^3-6x^2)} \\ x-1{\overline{\smash{\big)}\,x^3-6x^2+11x-6\phantom{)}}} \\ \underline{-~\phantom{(}(x^3-x^2)\phantom{-5x^2+11x-6)}} \\ -5x^2+11x \phantom{-6)} \\ \underline{-~\phantom{(}(-5x^2+5x)\phantom{-6)}} \\ 6x-6 \phantom{)} \\ \underline{-~\phantom{(}(6x-6)} \\ 0 \phantom{)} \end{array}$

Next, we factorise the quotient $x^2 - 5x + 6$:

$x^2 - 3x - 2x + 6$

$x(x - 3) - 2(x - 3)$

$(x - 3)(x - 2)$

Therefore, $x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)$.

Solution (iii): $x^3 + x^2 - 4x - 4$

Let $p(x) = x^3 + x^2 - 4x - 4$.

In this polynomial, we can group the terms directly to find the factors:

$x^2(x + 1) - 4(x + 1)$

Taking $(x + 1)$ as common:

$(x + 1)(x^2 - 4)$

$(x + 1)(x - 2)(x + 2)$

Therefore, $x^3 + x^2 - 4x - 4 = (x + 1)(x - 2)(x + 2)$.

Solution (iv): $3x^3 - x^2 - 3x + 1$

Let $p(x) = 3x^3 - x^2 - 3x + 1$.

Using the method of grouping terms:

$x^2(3x - 1) - 1(3x - 1)$

Taking $(3x - 1)$ as common:

$(3x - 1)(x^2 - 1)$

$(3x - 1)(x - 1)(x + 1)$

Therefore, $3x^3 - x^2 - 3x + 1 = (3x - 1)(x - 1)(x + 1)$.

Solution (i):

We need to evaluate $103^3$ using a suitable identity.

We can write $103$ as $100 + 3$. So, $103^3 = (100+3)^3$.

We use the identity: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Here, $a = 100$ and $b = 3$.

$(100+3)^3 = (100)^3 + 3(100)^2(3) + 3(100)(3)^2 + (3)^3$

$= 1000000 + 3(10000)(3) + 3(100)(9) + 27$

$= 1000000 + 90000 + 2700 + 27$

$= 1092727$

Thus, $103^3 = 1092727$.

Solution (ii):

We need to evaluate $101 \times 102$ using a suitable identity.

We can write $101$ as $100 + 1$ and $102$ as $100 + 2$.

So, $101 \times 102 = (100+1)(100+2)$.

We use the identity: $(x+a)(x+b) = x^2 + (a+b)x + ab$

Here, $x = 100$, $a = 1$, and $b = 2$.

$(100+1)(100+2) = (100)^2 + (1+2)(100) + (1)(2)$

$= 10000 + (3)(100) + 2$

$= 10000 + 300 + 2$

$= 10302$

Thus, $101 \times 102 = 10302$.

Solution (iii):

We need to evaluate $999^2$ using a suitable identity.

We can write $999$ as $1000 - 1$. So, $999^2 = (1000-1)^2$.

We use the identity: $(a-b)^2 = a^2 - 2ab + b^2$

Here, $a = 1000$ and $b = 1$.

$(1000-1)^2 = (1000)^2 - 2(1000)(1) + (1)^2$

$= 1000000 - 2000 + 1$

$= 998000 + 1$

$= 998001$

Thus, $999^2 = 998001$.

Solution (i):

We are asked to factorise $4x^2 + 20x + 25$.

We observe that the expression is in the form of a perfect square trinomial, $a^2 + 2ab + b^2$.

We can write $4x^2$ as $(2x)^2$ and $25$ as $(5)^2$.

So, we have $(2x)^2 + 20x + (5)^2$.

Let's check the middle term: $2 \times (2x) \times (5) = 20x$. This matches the given expression.

Using the identity: $a^2 + 2ab + b^2 = (a+b)^2$

Here, $a = 2x$ and $b = 5$.

Therefore, $4x^2 + 20x + 25 = (2x+5)^2$.

Solution (ii):

We are asked to factorise $9y^2 – 66yz + 121z^2$.

We observe that the expression is in the form of a perfect square trinomial, $a^2 - 2ab + b^2$.

We can write $9y^2$ as $(3y)^2$ and $121z^2$ as $(11z)^2$.

So, we have $(3y)^2 - 66yz + (11z)^2$.

Let's check the middle term: $2 \times (3y) \times (11z) = 66yz$. This matches the middle term with the correct sign.

Using the identity: $a^2 - 2ab + b^2 = (a-b)^2$

Here, $a = 3y$ and $b = 11z$.

Therefore, $9y^2 – 66yz + 121z^2 = (3y-11z)^2$.

Solution (iii):

We are asked to factorise $\left( 2x + \frac{1}{3} \right)^{2}-\left( x - \frac{1}{2} \right)^{2}$.

This expression is in the form $A^2 - B^2$, where $A = \left( 2x + \frac{1}{3} \right)$ and $B = \left( x - \frac{1}{2} \right)$.

Using the identity: $A^2 - B^2 = (A+B)(A-B)$

First, let's find $A+B$:

$A+B = \left( 2x + \frac{1}{3} \right) + \left( x - \frac{1}{2} \right)$

$A+B = 2x + \frac{1}{3} + x - \frac{1}{2}$

$A+B = (2x + x) + \left( \frac{1}{3} - \frac{1}{2} \right)$

$A+B = 3x + \left( \frac{2}{6} - \frac{3}{6} \right)$

$A+B = 3x - \frac{1}{6}$

Next, let's find $A-B$:

$A-B = \left( 2x + \frac{1}{3} \right) - \left( x - \frac{1}{2} \right)$

$A-B = 2x + \frac{1}{3} - x + \frac{1}{2}$

$A-B = (2x - x) + \left( \frac{1}{3} + \frac{1}{2} \right)$

$A-B = x + \left( \frac{2}{6} + \frac{3}{6} \right)$

$A-B = x + \frac{5}{6}$

Now, substitute $A+B$ and $A-B$ back into the identity $(A+B)(A-B)$:

$\left( 2x + \frac{1}{3} \right)^{2}-\left( x - \frac{1}{2} \right)^{2} = \left( 3x - \frac{1}{6} \right)\left( x + \frac{5}{6} \right)$

Solution (i):

We are asked to factorise $9x^2 – 12x + 3$.

First, we look for a common factor among the terms. The coefficients are 9, -12, and 3. The greatest common divisor is 3.

So, we can take out 3 as a common factor:

$9x^2 – 12x + 3 = 3(3x^2 – 4x + 1)$

Now, we need to factorise the quadratic expression $3x^2 – 4x + 1$. We can use the method of splitting the middle term.

We need to find two numbers whose product is the product of the coefficient of $x^2$ and the constant term, i.e., $3 \times 1 = 3$, and whose sum is the coefficient of the middle term, which is -4.

The two numbers are -3 and -1, because $(-3) \times (-1) = 3$ and $(-3) + (-1) = -4$.

We split the middle term $-4x$ into $-3x - x$:

$3x^2 – 4x + 1 = 3x^2 - 3x - x + 1$

Now, we group the terms and factor by grouping:

$(3x^2 - 3x) + (-x + 1)$

$3x(x - 1) - 1(x - 1)$

Now, take the common factor $(x-1)$:

$(x - 1)(3x - 1)$

Combining the common factor we initially took out, the factorization of $9x^2 – 12x + 3$ is:

$3(x - 1)(3x - 1)$

Solution (ii):

We are asked to factorise $9x^2 – 12x + 4$.

We observe that this expression might be a perfect square trinomial of the form $a^2 - 2ab + b^2$.

Let's examine the first and the last terms:

$9x^2 = (3x)^2$

$4 = (2)^2$

Now, let's check if the middle term $-12x$ matches $-2ab$ with $a=3x$ and $b=2$:

$-2ab = -2(3x)(2) = -12x$

This matches the middle term of the given expression.

Using the identity: $a^2 - 2ab + b^2 = (a-b)^2$

Here, $a = 3x$ and $b = 2$.

Therefore, $9x^2 – 12x + 4 = (3x - 2)^2$.

Solution (i):

We need to expand $(4a – b + 2c)^2$.

We use the identity: $(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

Here, $x = 4a$, $y = -b$, and $z = 2c$.

Substituting these values into the identity:

$(4a – b + 2c)^2 = (4a)^2 + (-b)^2 + (2c)^2 + 2(4a)(-b) + 2(-b)(2c) $$ + 2(2c)(4a)$

$= 16a^2 + b^2 + 4c^2 - 8ab - 4bc + 16ac$

Thus, the expansion is $16a^2 + b^2 + 4c^2 - 8ab - 4bc + 16ac$.

Solution (ii):

We need to expand $(3a – 5b – c)^2$.

We use the identity: $(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

Here, $x = 3a$, $y = -5b$, and $z = -c$.

Substituting these values into the identity:

$(3a – 5b – c)^2 = (3a)^2 + (-5b)^2 + (-c)^2 + 2(3a)(-5b) + 2(-5b)(-c) $$ + 2(-c)(3a)$

$= 9a^2 + 25b^2 + c^2 - 30ab + 10bc - 6ac$

Thus, the expansion is $9a^2 + 25b^2 + c^2 - 30ab + 10bc - 6ac$.

Solution (iii):

We need to expand $(– x + 2y – 3z)^2$.

We use the identity: $(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

Here, $x = -x$, $y = 2y$, and $z = -3z$.

Substituting these values into the identity:

$(– x + 2y – 3z)^2 = (-x)^2 + (2y)^2 + (-3z)^2 + 2(-x)(2y) $$ + 2(2y)(-3z) + 2(-3z)(-x)$

$= x^2 + 4y^2 + 9z^2 - 4xy - 12yz + 6xz$

Thus, the expansion is $x^2 + 4y^2 + 9z^2 - 4xy - 12yz + 6xz$.

To Factorise:

The given expressions are in the form of the algebraic identity:

Solution (i): $9x^2 + 4y^2 + 16z^2 + 12xy - 16yz - 24xz$

We can rewrite the terms as follows:

$9x^2 = (3x)^2$

$4y^2 = (2y)^2$

$16z^2 = (4z)^2$

Looking at the signs of the terms $12xy$, $-16yz$, and $-24xz$, we observe that the terms involving $z$ are negative. This implies that the term containing $z$ is negative.

Rewriting the expression in the standard identity form:

$= (3x)^2 + (2y)^2 + (-4z)^2 + 2(3x)(2y) + 2(2y)(-4z) + 2(-4z)(3x)$

By comparing this with the identity in equation (i), we have $a = 3x$, $b = 2y$, and $c = -4z$.

Hence, $9x^2 + 4y^2 + 16z^2 + 12xy - 16yz - 24xz = (3x + 2y - 4z)^2$ or $(3x + 2y - 4z)(3x + 2y - 4z)$.

Solution (ii): $25x^2 + 16y^2 + 4z^2 - 40xy + 16yz - 20xz$

We can rewrite the terms as follows:

$25x^2 = (5x)^2$

$16y^2 = (4y)^2$

$4z^2 = (2z)^2$

Observing the signs: $-40xy$ and $-20xz$ are negative. The common variable in these negative terms is $x$. This implies that the term containing $x$ is negative.

Rewriting the expression:

$= (-5x)^2 + (4y)^2 + (2z)^2 + 2(-5x)(4y) + 2(4y)(2z) + 2(2z)(-5x)$

Comparing with the identity, we have $a = -5x$, $b = 4y$, and $c = 2z$.

Hence, $25x^2 + 16y^2 + 4z^2 - 40xy + 16yz - 20xz = (-5x + 4y + 2z)^2$ or $(5x - 4y - 2z)(5x - 4y - 2z)$.

Solution (iii): $16x^2 + 4y^2 + 9z^2 - 16xy - 12yz + 24xz$

We can rewrite the terms as follows:

$16x^2 = (4x)^2$

$4y^2 = (2y)^2$

$9z^2 = (3z)^2$

Observing the signs: $-16xy$ and $-12yz$ are negative. The common variable in these negative terms is $y$. This implies that the term containing $y$ is negative.

Rewriting the expression:

$= (4x)^2 + (-2y)^2 + (3z)^2 + 2(4x)(-2y) + 2(-2y)(3z) + 2(3z)(4x)$

Comparing with the identity, we have $a = 4x$, $b = -2y$, and $c = 3z$.

Hence, $16x^2 + 4y^2 + 9z^2 - 16xy - 12yz + 24xz = (4x - 2y + 3z)^2$ or $(4x - 2y + 3z)(4x - 2y + 3z)$.

Given:

$a + b + c = 9$

$ab + bc + ca = 26$

To Find:

$a^2 + b^2 + c^2$

Solution:

We use the identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$

We are given $a+b+c = 9$ and $ab+bc+ca = 26$.

Substitute these values into the identity:

$(9)^2 = a^2 + b^2 + c^2 + 2(26)$

Calculate the square of 9 and the product of 2 and 26:

$81 = a^2 + b^2 + c^2 + 52$

Now, we need to isolate $a^2 + b^2 + c^2$. Subtract 52 from both sides of the equation:

$81 - 52 = a^2 + b^2 + c^2$

Calculate the difference:

$29 = a^2 + b^2 + c^2$

Therefore, $a^2 + b^2 + c^2 = 29$.

Solution (i):

We need to expand $(3a – 2b)^3$.

We use the identity for the cube of a binomial difference: $(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$

Here, $x = 3a$ and $y = 2b$.

Substituting these values into the identity:

$(3a – 2b)^3 = (3a)^3 - 3(3a)^2(2b) + 3(3a)(2b)^2 - (2b)^3$

$= 27a^3 - 3(9a^2)(2b) + 3(3a)(4b^2) - 8b^3$

$= 27a^3 - 54a^2b + 36ab^2 - 8b^3$

Thus, the expansion is $27a^3 - 54a^2b + 36ab^2 - 8b^3$.

Solution (ii):

We need to expand $\left( \frac{1}{x} + \frac{y}{3} \right)^{3}$.

We use the identity for the cube of a binomial sum: $(p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3$

Here, $p = \frac{1}{x}$ and $q = \frac{y}{3}$.

Substituting these values into the identity:

$\left( \frac{1}{x} + \frac{y}{3} \right)^{3} = \left(\frac{1}{x}\right)^3 + 3\left(\frac{1}{x}\right)^2\left(\frac{y}{3}\right) + 3\left(\frac{1}{x}\right)\left(\frac{y}{3}\right)^2 + \left(\frac{y}{3}\right)^3$

$= \frac{1}{x^3} + 3\left(\frac{1}{x^2}\right)\left(\frac{y}{3}\right) + 3\left(\frac{1}{x}\right)\left(\frac{y^2}{9}\right) + \frac{y^3}{27}$

$= \frac{1}{x^3} + \frac{3y}{3x^2} + \frac{3y^2}{9x} + \frac{y^3}{27}$

$= \frac{1}{x^3} + \frac{y}{x^2} + \frac{y^2}{3x} + \frac{y^3}{27}$

Thus, the expansion is $\frac{1}{x^3} + \frac{y}{x^2} + \frac{y^2}{3x} + \frac{y^3}{27}$.

Solution (iii):

We need to expand $\left( 4 - \frac{1}{3x} \right)^{3}$.

We use the identity for the cube of a binomial difference: $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

Here, $a = 4$ and $b = \frac{1}{3x}$.

Substituting these values into the identity:

$\left( 4 - \frac{1}{3x} \right)^{3} = (4)^3 - 3(4)^2\left(\frac{1}{3x}\right) + 3(4)\left(\frac{1}{3x}\right)^2 - \left(\frac{1}{3x}\right)^3$

$= 64 - 3(16)\left(\frac{1}{3x}\right) + 12\left(\frac{1}{9x^2}\right) - \frac{1}{27x^3}$

$= 64 - \frac{48}{3x} + \frac{12}{9x^2} - \frac{1}{27x^3}$

$= 64 - \frac{16}{x} + \frac{4}{3x^2} - \frac{1}{27x^3}$

Thus, the expansion is $64 - \frac{16}{x} + \frac{4}{3x^2} - \frac{1}{27x^3}$.

Solution (i):

We are asked to factorise $1 – 64a^3 – 12a + 48a^2$.

Let's rearrange the terms: $1 + 48a^2 - 12a - 64a^3$.

We observe that this expression involves cubic terms and constants, which suggests using the identity for the cube of a binomial difference: $(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$.

We can write $1$ as $1^3$ and $64a^3$ as $(4a)^3$. The term $-64a^3$ can be written as $(-4a)^3$.

Let's consider if the expression is of the form $(1 - 4a)^3$.

Using the identity with $x=1$ and $y=4a$:

$(1 - 4a)^3 = (1)^3 - 3(1)^2(4a) + 3(1)(4a)^2 - (4a)^3$

$= 1 - 3(1)(4a) + 3(1)(16a^2) - 64a^3$

$= 1 - 12a + 48a^2 - 64a^3$

Rearranging the terms, we get $1 - 64a^3 - 12a + 48a^2$, which matches the given expression.

Therefore, the factorization is $(1 - 4a)^3$ or $(1 - 4a)(1 - 4a)(1 - 4a)$.

Solution (ii):

We are asked to factorise $8p^3 + \frac{12}{5}p^2 + \frac{6}{25}p + \frac{1}{125}$.

We observe that this expression involves cubic terms and constants, which suggests using the identity for the cube of a binomial sum: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

We can write the first term $8p^3$ as $(2p)^3$. So, we can take $a = 2p$.

We can write the last term $\frac{1}{125}$ as $\left(\frac{1}{5}\right)^3$. So, we can take $b = \frac{1}{5}$.

Now, let's check if the middle terms match the expansion of $(2p + \frac{1}{5})^3$:

$3a^2b = 3(2p)^2\left(\frac{1}{5}\right) = 3(4p^2)\left(\frac{1}{5}\right) = \frac{12}{5}p^2$. This matches the second term.

$3ab^2 = 3(2p)\left(\frac{1}{5}\right)^2 = 3(2p)\left(\frac{1}{25}\right) = \frac{6}{25}p$. This matches the third term.

Since all terms match the expansion of $(a+b)^3$ with $a=2p$ and $b=\frac{1}{5}$, the given expression is a perfect cube.

Therefore, the factorization is $\left(2p + \frac{1}{5}\right)^3$ or $\left(2p + \frac{1}{5}\right) \left(2p + \frac{1}{5}\right) \left(2p + \frac{1}{5}\right)$.

Solution (i):

We need to find the product $\left( \frac{x}{2} + 2y \right)\left( \frac{x^2}{4} - xy + 4y^2 \right)$.

We observe that this expression is in the form $(a+b)(a^2 - ab + b^2)$, which is the expansion of the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.

Let $a = \frac{x}{2}$ and $b = 2y$.

Then, $a^2 = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}$, $b^2 = (2y)^2 = 4y^2$, and $ab = \left(\frac{x}{2}\right)(2y) = xy$.

The second factor is $\frac{x^2}{4} - xy + 4y^2$, which is $a^2 - ab + b^2$.

So, the product is $a^3 + b^3$ with $a = \frac{x}{2}$ and $b = 2y$.

$\left( \frac{x}{2} + 2y \right)\left( \frac{x^2}{4} - xy + 4y^2 \right) = \left(\frac{x}{2}\right)^3 + (2y)^3$

$= \frac{x^3}{8} + 8y^3$

Thus, the product is $\frac{x^3}{8} + 8y^3$.

Solution (ii):

We need to find the product $(x^2 – 1) (x^4 + x^2 + 1)$.

We observe that this expression is in the form $(a-b)(a^2 + ab + b^2)$, which is the expansion of the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

Let $a = x^2$ and $b = 1$.

Then, $a^2 = (x^2)^2 = x^4$, $b^2 = 1^2 = 1$, and $ab = (x^2)(1) = x^2$.

The second factor is $x^4 + x^2 + 1$, which is $a^2 + ab + b^2$.

So, the product is $a^3 - b^3$ with $a = x^2$ and $b = 1$.

$(x^2 – 1) (x^4 + x^2 + 1) = (x^2)^3 - (1)^3$

$= x^{6} - 1$

Thus, the product is $x^6 - 1$.

Solution (i):

We are asked to factorise $1 + 64x^3$.

We observe that this expression is a sum of two cubes. We can write $1$ as $1^3$ and $64x^3$ as $(4x)^3$.

So, the expression is in the form $a^3 + b^3$, where $a = 1$ and $b = 4x$.

We use the identity: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$

Substitute $a = 1$ and $b = 4x$ into the identity:

$1^3 + (4x)^3 = (1 + 4x)((1)^2 - (1)(4x) + (4x)^2)$

$= (1 + 4x)(1 - 4x + 16x^2)$

Thus, the factorization is $(1 + 4x)(1 - 4x + 16x^2)$.

Solution (ii):

We are asked to factorise $a^3 – 2\sqrt{2}b^3$.

We observe that this expression is a difference of two cubes. We can write $a^3$ as $(a)^3$.

For the second term, we notice that $2\sqrt{2} = (\sqrt{2})^2 \times \sqrt{2} = (\sqrt{2})^3$.

So, $2\sqrt{2}b^3 = (\sqrt{2})^3 b^3 = (\sqrt{2}b)^3$.

The expression is in the form $x^3 - y^3$, where $x = a$ and $y = \sqrt{2}b$.

We use the identity: $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$

Substitute $x = a$ and $y = \sqrt{2}b$ into the identity:

$a^3 - (\sqrt{2}b)^3 = (a - \sqrt{2}b)(a^2 + a(\sqrt{2}b) + (\sqrt{2}b)^2)$

$= (a - \sqrt{2}b)(a^2 + \sqrt{2}ab + 2b^2)$

Thus, the factorization is $(a - \sqrt{2}b)(a^2 + \sqrt{2}ab + 2b^2)$.

We need to find the product of $(2x – y + 3z)$ and $(4x^2 + y^2 + 9z^2 + 2xy + 3yz – 6xz)$.

We observe that this product is in the form $(a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

Let $a = 2x$, $b = -y$, and $c = 3z$.

Let's check if the second factor matches $a^2 + b^2 + c^2 - ab - bc - ca$ with these values:

$a^2 = (2x)^2 = 4x^2$

$b^2 = (-y)^2 = y^2$

$c^2 = (3z)^2 = 9z^2$

$-ab = -(2x)(-y) = 2xy$

$-bc = -(-y)(3z) = 3yz$

$-ca = -(3z)(2x) = -6xz$

The second factor is $4x^2 + y^2 + 9z^2 + 2xy + 3yz - 6xz$, which exactly matches $a^2 + b^2 + c^2 - ab - bc - ca$ with $a=2x$, $b=-y$, and $c=3z$.

The first factor is $2x - y + 3z$, which is $a+b+c$.

We use the identity: $(a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc$

Substituting $a = 2x$, $b = -y$, and $c = 3z$ into the right side of the identity:

$(2x)^3 + (-y)^3 + (3z)^3 - 3(2x)(-y)(3z)$

$= 8x^3 - y^3 + 27z^3 - 3(-6xyz)$

$= 8x^3 - y^3 + 27z^3 + 18xyz$

Thus, the product is $8x^3 - y^3 + 27z^3 + 18xyz$.

Solution (i):

We are asked to factorise $a^3 – 8b^3 – 64c^3 – 24abc$.

We can rewrite the terms as cubes:

$a^3 = (a)^3$

$-8b^3 = (-2b)^3$

$-64c^3 = (-4c)^3$

The expression can be written as $(a)^3 + (-2b)^3 + (-4c)^3 - 24abc$.

This expression is in the form $x^3 + y^3 + z^3 - 3xyz$, where $x = a$, $y = -2b$, and $z = -4c$.

Let's check the $-3xyz$ term:

$-3xyz = -3(a)(-2b)(-4c) = -3(8abc) = -24abc$. This matches the given term.

We use the identity: $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Substitute $x = a$, $y = -2b$, and $z = -4c$ into the identity:

$(a + (-2b) + (-4c))((a)^2 + (-2b)^2 + (-4c)^2 - (a)(-2b) $$ - (-2b)(-4c) - (-4c)(a))$

$= (a - 2b - 4c)(a^2 + 4b^2 + 16c^2 + 2ab - 8bc + 4ca)$

Thus, the factorization is $(a - 2b - 4c)(a^2 + 4b^2 + 16c^2 + 2ab - 8bc + 4ca)$.

Solution (ii):

We are asked to factorise $2\sqrt{2}a^3 + 8b^3 – 27c^3 + 18\sqrt{2}abc$.

We can rewrite the terms as cubes:

$2\sqrt{2}a^3 = (\sqrt{2})^3 a^3 = (\sqrt{2}a)^3$

$8b^3 = (2b)^3$

$-27c^3 = (-3c)^3$

The expression can be written as $(\sqrt{2}a)^3 + (2b)^3 + (-3c)^3 + 18\sqrt{2}abc$.

This expression is in the form $x^3 + y^3 + z^3 - 3xyz$, where $x = \sqrt{2}a$, $y = 2b$, and $z = -3c$.

Let's check the $-3xyz$ term:

$-3xyz = -3(\sqrt{2}a)(2b)(-3c) = -3(\sqrt{2}a)(-6bc) = 18\sqrt{2}abc$. This matches the given term.

We use the identity: $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Substitute $x = \sqrt{2}a$, $y = 2b$, and $z = -3c$ into the identity:

$(\sqrt{2}a + 2b + (-3c))((\sqrt{2}a)^2 + (2b)^2 + (-3c)^2 - (\sqrt{2}a)(2b) $$ - (2b)(-3c) - (-3c)(\sqrt{2}a))$

$= (\sqrt{2}a + 2b - 3c)(2a^2 + 4b^2 + 9c^2 - 2\sqrt{2}ab + 6bc + 3\sqrt{2}ca)$

Thus, the factorization is $(\sqrt{2}a + 2b - 3c)(2a^2 + 4b^2 + 9c^2 - 2\sqrt{2}ab + 6bc + 3\sqrt{2}ca)$.

Solution (i):

We need to find the value of $\left( \frac{1}{2} \right)^{3}+\left( \frac{1}{3} \right)^{3}-\left( \frac{5}{6} \right)^{3}$ without calculating the cubes.

We can rewrite the expression as $\left( \frac{1}{2} \right)^{3}+\left( \frac{1}{3} \right)^{3}+\left( -\frac{5}{6} \right)^{3}$.

Let $a = \frac{1}{2}$, $b = \frac{1}{3}$, and $c = -\frac{5}{6}$.

Let's check the sum $a+b+c$:

$a+b+c = \frac{1}{2} + \frac{1}{3} + \left(-\frac{5}{6}\right) = \frac{3}{6} + \frac{2}{6} - \frac{5}{6} = \frac{3+2-5}{6} = \frac{0}{6} = 0$

Since $a+b+c = 0$, we can use the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Using this identity, the value of the expression is $3abc$:

Value $= 3 \times \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times \left(-\frac{5}{6}\right)$

Value $= \cancel{3} \times \frac{1}{2} \times \frac{1}{\cancel{3}} \times \left(-\frac{5}{6}\right)$

Value $= \frac{1}{2} \times 1 \times \left(-\frac{5}{6}\right) = -\frac{5}{12}$

Thus, $\left( \frac{1}{2} \right)^{3}+\left( \frac{1}{3} \right)^{3}-\left( \frac{5}{6} \right)^{3} = -\frac{5}{12}$.

Solution (ii):

We need to find the value of $(0.2)^3 – (0.3)^3 + (0.1)^3$ without calculating the cubes.

We can rewrite the expression as $(0.2)^3 + (-0.3)^3 + (0.1)^3$.

Let $a = 0.2$, $b = -0.3$, and $c = 0.1$.

Let's check the sum $a+b+c$:

$a+b+c = 0.2 + (-0.3) + 0.1 = 0.2 - 0.3 + 0.1 = 0.3 - 0.3 = 0$

Since $a+b+c = 0$, we can use the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Using this identity, the value of the expression is $3abc$:

Value $= 3 \times (0.2) \times (-0.3) \times (0.1)$

Value $= 3 \times (0.02 \times -0.3)$

Value $= 3 \times (-0.006)$

Value $= -0.018$

Thus, $(0.2)^3 – (0.3)^3 + (0.1)^3 = -0.018$.

We are asked to factorise $(x – 2y)^3 + (2y – 3z)^3 + (3z – x)^3$ without finding the cubes.

Let $a = x – 2y$, $b = 2y – 3z$, and $c = 3z – x$.

The given expression is in the form $a^3 + b^3 + c^3$.

Let's find the sum of $a$, $b$, and $c$:

$a+b+c = (x – 2y) + (2y – 3z) + (3z – x)$

$a+b+c = x - 2y + 2y - 3z + 3z - x$

$a+b+c = (x-x) + (-2y+2y) + (-3z+3z)$

$a+b+c = 0 + 0 + 0 = 0$

Since $a+b+c = 0$, we can use the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Using this identity with $a = x – 2y$, $b = 2y – 3z$, and $c = 3z – x$, the factorization of the given expression is $3abc$:

$(x – 2y)^3 + (2y – 3z)^3 + (3z – x)^3 = 3(x – 2y)(2y – 3z)(3z – x)$

Thus, the factorization is $3(x – 2y)(2y – 3z)(3z – x)$.

Solution (i):

We need to find the value of $x^3 + y^3 – 12xy + 64$, when $x + y = – 4$.

The given expression is $x^3 + y^3 – 12xy + 64$.

We can rewrite the expression as $x^3 + y^3 + 4^3 - 12xy$.

This expression is in the form $a^3 + b^3 + c^3 - 3abc$, where $a=x$, $b=y$, and $c=4$.

Let's check the term $-3abc$:

$-3abc = -3(x)(y)(4) = -12xy$. This matches the term in the given expression.

We are given the condition $x + y = -4$.

Let's check the sum $a+b+c$:

$a+b+c = x + y + 4$

Substitute the given condition $x+y = -4$:

$a+b+c = (-4) + 4 = 0$

We use the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 - 3abc = 0$.

In our case, $a=x$, $b=y$, $c=4$. Since $a+b+c=0$, the value of the expression $a^3 + b^3 + c^3 - 3abc$ is 0.

Therefore, the value of $x^3 + y^3 + 4^3 - 3(x)(y)(4)$ is 0.

So, $x^3 + y^3 – 12xy + 64 = 0$.

Solution (ii):

We need to find the value of $x^3 – 8y^3 – 36xy – 216$, when $x = 2y + 6$.

The given condition is $x = 2y + 6$. We can rearrange this as $x - 2y - 6 = 0$.

The given expression is $x^3 – 8y^3 – 36xy – 216$.

We can rewrite the expression by grouping terms and considering signs for potential cube roots:

$x^3 + (-8y^3) + (-216) - 36xy$

We can write the first three terms as cubes:

$x^3 = (x)^3$

$-8y^3 = (-2y)^3$

$-216 = (-6)^3$

So the expression is $(x)^3 + (-2y)^3 + (-6)^3 - 36xy$.

This expression is in the form $a^3 + b^3 + c^3 - 3abc$, where $a=x$, $b=-2y$, and $c=-6$.

Let's check the term $-3abc$:

$-3abc = -3(x)(-2y)(-6) = -3(12xy) = -36xy$. This matches the term in the given expression.

Now let's check the sum $a+b+c$:

$a+b+c = x + (-2y) + (-6) = x - 2y - 6$

From the given condition $x = 2y + 6$, we have $x - 2y - 6 = 0$.

So, $a+b+c = 0$.

We use the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 - 3abc = 0$.

In our case, $a=x$, $b=-2y$, $c=-6$. Since $a+b+c=0$, the value of the expression $a^3 + b^3 + c^3 - 3abc$ is 0.

Therefore, the value of $(x)^3 + (-2y)^3 + (-6)^3 - 3(x)(-2y)(-6)$ is 0.

So, $x^3 – 8y^3 – 36xy – 216 = 0$.

The area of the rectangle is given by the expression $4a^2 + 4a – 3$.

The area of a rectangle is the product of its length and breadth. To find the possible expressions for the length and breadth, we need to factorise the given quadratic expression.

We will factorise the expression $4a^2 + 4a – 3$ by splitting the middle term.

We look for two numbers whose product is $(4) \times (-3) = -12$ and whose sum is the coefficient of the middle term, which is 4.

The two numbers are 6 and -2, since $6 \times (-2) = -12$ and $6 + (-2) = 4$.

Now, we split the middle term $4a$ as $6a - 2a$:

$4a^2 + 4a – 3 = 4a^2 + 6a - 2a – 3$

Group the terms:

$= (4a^2 + 6a) + (-2a – 3)$

Factor out common factors from each group:

$= 2a(2a + 3) - 1(2a + 3)$

Factor out the common binomial factor $(2a+3)$:

$= (2a + 3)(2a - 1)$

The factorization of the area is $(2a + 3)(2a - 1)$.

Since Area = Length $\times$ Breadth, the possible expressions for the length and breadth are the two factors.

Possible expression for Length = $2a + 3$

Possible expression for Breadth = $2a - 1$

(Alternatively, Length = $2a - 1$ and Breadth = $2a + 3$. We assume the dimension with the potentially larger value is the length, but without knowing the value of 'a', either expression can be length or breadth).

Given:

$x + y = 12$

$xy = 27$

To Find:

The value of $x^3 + y^3$.

Solution:

We use the identity for the sum of cubes: $x^3 + y^3 = (x+y)^3 - 3xy(x+y)$

We are given the values of $(x+y)$ and $xy$.

Substitute the given values into the identity:

$x^3 + y^3 = (12)^3 - 3(27)(12)$

First, calculate $(12)^3$:

$(12)^3 = 12 \times 12 \times 12 = 144 \times 12$

$(12)^3 = 1728$

Next, calculate $3(27)(12)$:

$3 \times 27 = 81$

$3(27)(12) = 81 \times 12$

$81 \times 12 = 972$

Now, substitute these values back into the equation for $x^3 + y^3$:

$x^3 + y^3 = 1728 - 972$

$x^3 + y^3 = 756$

Thus, the value of $x^3 + y^3$ is 756.

Given:

Polynomials: $P(z) = az^3 + 4z^2 + 3z – 4$ and $Q(z) = z^3 – 4z + a$.

When divided by $z – 3$, both polynomials leave the same remainder.

To Find:

The value of $a$.

Solution:

According to the Remainder Theorem, when a polynomial $P(x)$ is divided by $(x-c)$, the remainder is $P(c)$.

In this case, the divisor is $z-3$, so $c=3$.

Let $R_1$ be the remainder when $P(z)$ is divided by $z-3$.

$R_1 = P(3)$

$P(3) = a(3)^3 + 4(3)^2 + 3(3) - 4$

$P(3) = a(27) + 4(9) + 9 - 4$

$P(3) = 27a + 36 + 9 - 4$

$R_1 = 27a + 41$

Let $R_2$ be the remainder when $Q(z)$ is divided by $z-3$.

$R_2 = Q(3)$

$Q(3) = (3)^3 - 4(3) + a$

$Q(3) = 27 - 12 + a$

$R_2 = 15 + a$

We are given that the remainders are the same, so $R_1 = R_2$.

$27a + 41 = 15 + a$

Now, we solve this linear equation for $a$.

Subtract $a$ from both sides:

$27a - a + 41 = 15$

$26a + 41 = 15$

Subtract 41 from both sides:

$26a = 15 - 41$

$26a = -26$

Divide both sides by 26:

$a = \frac{-26}{26}$

$a = -1$

Thus, the value of $a$ is -1.

Given:

Polynomial: $p(x) = x^4 – 2x^3 + 3x^2 – ax + 3a – 7$.

When $p(x)$ is divided by $x + 1$, the remainder is 19.

To Find:

The value of $a$.

The remainder when $p(x)$ is divided by $x + 2$.

Solution:

According to the Remainder Theorem, when a polynomial $p(x)$ is divided by $(x-c)$, the remainder is $p(c)$.

For the first part, the divisor is $x+1$. This can be written as $x - (-1)$. So, $c=-1$. The remainder is given as 19.

Using the Remainder Theorem, the remainder is $p(-1)$.

Now, substitute $x = -1$ into the polynomial $p(x) = x^4 – 2x^3 + 3x^2 – ax + 3a – 7$:

$p(-1) = (-1)^4 – 2(-1)^3 + 3(-1)^2 – a(-1) + 3a – 7$

Calculate the terms:

$(-1)^4 = 1$

$(-1)^3 = -1$

$(-1)^2 = 1$

Substitute these values back into $p(-1)$:

$p(-1) = 1 – 2(-1) + 3(1) – a(-1) + 3a – 7$

$p(-1) = 1 + 2 + 3 + a + 3a – 7$

Combine the constant terms and the terms with $a$:

$p(-1) = (1 + 2 + 3 – 7) + (a + 3a)$

$p(-1) = (6 – 7) + 4a$

$p(-1) = -1 + 4a$

From equation (i), we have $p(-1) = 19$. So,

$4a - 1 = 19$

Now, solve for $a$:

$4a = 19 + 1$

$4a = 20$

$a = \frac{20}{4}$

$a = 5$

The value of $a$ is 5.

Now we need to find the remainder when $p(x)$ is divided by $x+2$.

First, substitute the value $a=5$ into the polynomial $p(x)$:

$p(x) = x^4 – 2x^3 + 3x^2 – (5)x + 3(5) – 7$

$p(x) = x^4 – 2x^3 + 3x^2 – 5x + 15 – 7$

$p(x) = x^4 – 2x^3 + 3x^2 – 5x + 8$

For the second part, the divisor is $x+2$, which means $c=-2$. The remainder is $p(-2)$.

Substitute $x = -2$ into the polynomial $p(x)$:

$p(-2) = (-2)^4 – 2(-2)^3 + 3(-2)^2 – 5(-2) + 8$

Calculate the terms:

$(-2)^4 = 16$

$(-2)^3 = -8$

$(-2)^2 = 4$

Substitute these values back into $p(-2)$:

$p(-2) = 16 – 2(-8) + 3(4) – 5(-2) + 8$

$p(-2) = 16 + 16 + 12 + 10 + 8$

Sum the terms:

$p(-2) = 32 + 12 + 10 + 8$

$p(-2) = 44 + 10 + 8$

$p(-2) = 54 + 8$

$p(-2) = 62$

The remainder when $p(x)$ is divided by $x+2$ is 62.

Given:

The polynomial is $P(x) = px^2 + 5x + r$.

$(x – 2)$ and $(x – \frac{1}{2})$ are factors of the polynomial $P(x)$.

To Show:

$p = r$

Solution:

According to the Factor Theorem, if $(x-c)$ is a factor of a polynomial $P(x)$, then $P(c) = 0$.

Since $(x – 2)$ is a factor of $P(x)$, we have $P(2) = 0$.

Substitute $x = 2$ into the polynomial $P(x) = px^2 + 5x + r$:

$P(2) = p(2)^2 + 5(2) + r$

$P(2) = 4p + 10 + r$

Since $P(2) = 0$, we have:

This can be rearranged as:

Since $(x – \frac{1}{2})$ is a factor of $P(x)$, we have $P(\frac{1}{2}) = 0$.

Substitute $x = \frac{1}{2}$ into the polynomial $P(x) = px^2 + 5x + r$:

$P\left(\frac{1}{2}\right) = p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r$

$P\left(\frac{1}{2}\right) = p\left(\frac{1}{4}\right) + \frac{5}{2} + r$

$P\left(\frac{1}{2}\right) = \frac{p}{4} + \frac{5}{2} + r$

Since $P(\frac{1}{2}) = 0$, we have:

To eliminate the fractions, multiply the entire equation by 4:

$4 \times \left(\frac{p}{4}\right) + 4 \times \left(\frac{5}{2}\right) + 4 \times r = 4 \times 0$

$p + 10 + 4r = 0$

This can be rearranged as:

Now we have a system of two linear equations:

Equation (1): $4p + r = -10$

Equation (2): $p + 4r = -10$

From Equation (1), we can express $r$ in terms of $p$:

Substitute the expression for $r$ from Equation (3) into Equation (2):

$p + 4(-10 - 4p) = -10$

$p - 40 - 16p = -10$

Combine like terms:

$-15p - 40 = -10$

Add 40 to both sides:

$-15p = -10 + 40$

$-15p = 30$

Divide both sides by -15:

$p = \frac{30}{-15}$

$p = -2$

Now substitute the value of $p = -2$ into Equation (3) to find the value of $r$:

$r = -10 - 4(-2)$

$r = -10 + 8$

$r = -2$

We found that $p = -2$ and $r = -2$.

Therefore, $p = r$.

Hence, it is shown that $p = r$.

Given:

Polynomial: $P(x) = 2x^4 – 5x^3 + 2x^2 – x + 2$.

Divisor: $D(x) = x^2 – 3x + 2$.

To Prove:

$P(x)$ is divisible by $D(x)$ without actual division.

Proof:

For $P(x)$ to be divisible by $D(x) = x^2 – 3x + 2$, $D(x)$ must be a factor of $P(x)$.

First, we factorise the divisor $D(x) = x^2 – 3x + 2$.

We find two numbers whose product is $(1) \times (2) = 2$ and whose sum is $-3$. These numbers are $-1$ and $-2$.

$x^2 – 3x + 2 = x^2 - x - 2x + 2$

$= x(x-1) - 2(x-1)$

$= (x-1)(x-2)$

So, the divisor $x^2 – 3x + 2$ is the product of two linear factors: $(x-1)$ and $(x-2)$.

For $P(x)$ to be divisible by $(x-1)(x-2)$, $P(x)$ must be divisible by each of these factors individually.

According to the Factor Theorem, a polynomial $P(x)$ is divisible by $(x-c)$ if and only if $P(c) = 0$.

We need to check if $P(1) = 0$ and $P(2) = 0$.

Evaluate $P(1)$ by substituting $x=1$ into $P(x) = 2x^4 – 5x^3 + 2x^2 – x + 2$:

$P(1) = 2(1)^4 – 5(1)^3 + 2(1)^2 – (1) + 2$

$P(1) = 2(1) – 5(1) + 2(1) – 1 + 2$

$P(1) = 2 – 5 + 2 – 1 + 2$

$P(1) = (2+2+2) - (5+1) = 6 - 6 = 0$

Since $P(1) = 0$, by the Factor Theorem, $(x-1)$ is a factor of $P(x)$.

Evaluate $P(2)$ by substituting $x=2$ into $P(x) = 2x^4 – 5x^3 + 2x^2 – x + 2$:

$P(2) = 2(2)^4 – 5(2)^3 + 2(2)^2 – (2) + 2$

$P(2) = 2(16) – 5(8) + 2(4) – 2 + 2$

$P(2) = 32 – 40 + 8 – 2 + 2$

$P(2) = (32+8+2) - (40+2) = 42 - 42 = 0$

Since $P(2) = 0$, by the Factor Theorem, $(x-2)$ is a factor of $P(x)$.

Since $P(x)$ is divisible by both $(x-1)$ and $(x-2)$, and these are distinct linear factors, $P(x)$ is divisible by their product $(x-1)(x-2) = x^2 – 3x + 2$.

Therefore, $2x^4 – 5x^3 + 2x^2 – x + 2$ is divisible by $x^2 – 3x + 2$.

We need to simplify $(2x – 5y)^3 – (2x + 5y)^3$.

Let $a = 2x$ and $b = 5y$. The expression can be written as $(a-b)^3 - (a+b)^3$.

We use the identities for the cube of a binomial:

• $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
• $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Now, subtract the second expansion from the first:

$(a-b)^3 - (a+b)^3 = (a^3 - 3a^2b + 3ab^2 - b^3) $$ - (a^3 + 3a^2b + 3ab^2 + b^3)$

$= a^3 - 3a^2b + 3ab^2 - b^3 - a^3 - 3a^2b - 3ab^2 - b^3$

Combine like terms:

$= (a^3 - a^3) + (-3a^2b - 3a^2b) + (3ab^2 - 3ab^2) + (-b^3 - b^3)$

$= 0 - 6a^2b + 0 - 2b^3$

$= -6a^2b - 2b^3$

Now, substitute back $a = 2x$ and $b = 5y$ into the simplified expression:

$-6a^2b - 2b^3 = -6(2x)^2(5y) - 2(5y)^3$

$= -6(4x^2)(5y) - 2(125y^3)$

$= -6(20x^2y) - 250y^3$

$= -120x^2y - 250y^3$

Thus, the simplified expression is $-120x^2y - 250y^3$.

We need to multiply the expression $(x^2 + 4y^2 + z^2 + 2xy + xz – 2yz)$ by $(– z + x – 2y)$.

Let's rearrange the terms in the second factor: $(x - 2y - z)$.

The first factor is $x^2 + 4y^2 + z^2 + 2xy + xz – 2yz$.

Let's compare the given expressions with the identity for the sum of cubes minus 3 times the product: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

Let's try to identify $a$, $b$, and $c$ from the second factor $(x - 2y - z)$.

Let $a = x$, $b = -2y$, and $c = -z$.

Then $a+b+c = x + (-2y) + (-z) = x - 2y - z$, which matches the second given factor.

Now, let's calculate the terms in the second part of the identity $(a^2 + b^2 + c^2 - ab - bc - ca)$ using $a=x$, $b=-2y$, and $c=-z$:

$a^2 = (x)^2 = x^2$

$b^2 = (-2y)^2 = 4y^2$

$c^2 = (-z)^2 = z^2$

$-ab = -(x)(-2y) = 2xy$

$-bc = -(-2y)(-z) = -(2yz) = -2yz$

$-ca = -(-z)(x) = xz$

So, $a^2 + b^2 + c^2 - ab - bc - ca = x^2 + 4y^2 + z^2 + 2xy - 2yz + xz$.

This matches the first given factor, $x^2 + 4y^2 + z^2 + 2xy + xz – 2yz$.

Therefore, the product of the two given expressions is equal to $a^3 + b^3 + c^3 - 3abc$, with $a=x$, $b=-2y$, and $c=-z$.

Let's calculate $a^3$, $b^3$, $c^3$, and $3abc$:

$a^3 = (x)^3 = x^3$

$b^3 = (-2y)^3 = -8y^3$

$c^3 = (-z)^3 = -z^3$

$3abc = 3(x)(-2y)(-z) = 6xyz$

The product is $a^3 + b^3 + c^3 - 3abc$:

Product $= x^3 + (-8y^3) + (-z^3) - 6xyz$

Product $= x^3 - 8y^3 - z^3 - 6xyz$

Thus, the product is $x^3 - 8y^3 - z^3 - 6xyz$.

Given:

$a, b, c$ are all non–zero.

$a + b + c = 0$

To Prove:

$\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3$

Proof:

Consider the left-hand side (LHS) of the equation we need to prove:

LHS $= \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$

To add these fractions, we find a common denominator, which is the least common multiple of $bc, ca, ab$. The LCM is $abc$.

We rewrite each term with the common denominator $abc$:

$\frac{a^2}{bc} = \frac{a^2 \times a}{bc \times a} = \frac{a^3}{abc}$

$\frac{b^2}{ca} = \frac{b^2 \times b}{ca \times b} = \frac{b^3}{abc}$

$\frac{c^2}{ab} = \frac{c^2 \times c}{ab \times c} = \frac{c^3}{abc}$

Now, substitute these back into the expression for the LHS:

LHS $= \frac{a^3}{abc} + \frac{b^3}{abc} + \frac{c^3}{abc}$

LHS $= \frac{a^3 + b^3 + c^3}{abc}$

We are given that $a + b + c = 0$.

We know the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Using this identity, we can substitute $a^3 + b^3 + c^3$ in the numerator of the LHS expression:

LHS $= \frac{3abc}{abc}$

Since $a, b, c$ are all non-zero, their product $abc$ is also non-zero. Thus, we can cancel $abc$ from the numerator and the denominator.

LHS $= 3$

This is equal to the right-hand side (RHS) of the equation.

LHS $= 3 =$ RHS

Therefore, it is proven that $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3$ when $a+b+c=0$ and $a, b, c$ are non-zero.

Given:

$a + b + c = 5$

$ab + bc + ca = 10$

To Prove:

$a^3 + b^3 + c^3 – 3abc = – 25$

Proof:

We use the identity: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$

We are given $a+b+c = 5$ and $ab+bc+ca = 10$. However, we need the value of $a^2 + b^2 + c^2$.

We use another identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$

Substitute the given values into this identity:

$(5)^2 = a^2 + b^2 + c^2 + 2(10)$

$25 = a^2 + b^2 + c^2 + 20$

Now, solve for $a^2 + b^2 + c^2$:

$a^2 + b^2 + c^2 = 25 - 20$

$a^2 + b^2 + c^2 = 5$

Now we have all the required components for the main identity: $a+b+c = 5$, $a^2+b^2+c^2 = 5$, and $ab+bc+ca = 10$.

Substitute these values into the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$:

$a^3 + b^3 + c^3 - 3abc = (5)(5 - 10)$

$a^3 + b^3 + c^3 - 3abc = (5)(-5)$

$a^3 + b^3 + c^3 - 3abc = -25$

Thus, it is proven that $a^3 + b^3 + c^3 – 3abc = – 25$.

To Prove:

$(a + b + c)^3 – a^3 – b^3 – c^3 = 3(a + b ) (b + c) (c + a)$

Proof:

Consider the Left Hand Side (LHS) of the identity:

LHS $= (a + b + c)^3 – a^3 – b^3 – c^3$

We can group the terms and use the identity for the difference of cubes, $X^3 - Y^3 = (X-Y)(X^2+XY+Y^2)$.

Let $X = a+b+c$ and $Y = c$. Then the first part of the LHS is $X^3 - Y^3$.

$(a+b+c)^3 - c^3 = ((a+b+c) - c)((a+b+c)^2 $$ + (a+b+c)c + c^2)$

$= (a+b)((a+b)^2 + c^2 + 2(a+b)c + (a+b)c + c^2 + c^2)$

$= (a+b)((a+b)^2 + 3(a+b)c + 3c^2)$

$= (a+b)(a^2 + 2ab + b^2 + 3ac + 3bc + 3c^2)$

Now substitute this back into the LHS expression:

LHS $= (a+b)(a^2 + 2ab + b^2 + 3ac + 3bc + 3c^2) - a^3 - b^3$

We know the identity for the sum of cubes, $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.

Substitute this into the LHS expression:

LHS $= (a+b)(a^2 + 2ab + b^2 + 3ac + 3bc + 3c^2) - (a+b)(a^2 - ab + b^2)$

Factor out the common term $(a+b)$:

LHS $= (a+b)[ (a^2 + 2ab + b^2 + 3ac + 3bc + 3c^2) - (a^2 - ab + b^2) ]$

Remove the parenthesis inside the square bracket, changing the signs of the terms from the second group:

LHS $= (a+b)[ a^2 + 2ab + b^2 + 3ac + 3bc + 3c^2 - a^2 + ab - b^2 ]$

Combine like terms inside the square bracket ($a^2$ with $-a^2$, $b^2$ with $-b^2$, $2ab$ with $ab$):

LHS $= (a+b)[ (a^2 - a^2) + (b^2 - b^2) + (2ab + ab) + 3ac + 3bc + 3c^2 ]$

LHS $= (a+b)[ 0 + 0 + 3ab + 3ac + 3bc + 3c^2 ]$

LHS $= (a+b)(3ab + 3ac + 3bc + 3c^2)$

Factor out 3 from the terms inside the second parenthesis:

LHS $= (a+b)[ 3(ab + ac + bc + c^2) ]$

LHS $= 3(a+b)(ab + ac + bc + c^2)$

Now, factor the expression inside the second parenthesis by grouping the terms:

$ab + ac + bc + c^2 = (ab + ac) + (bc + c^2)$

$= a(b+c) + c(b+c)$

$= (a+c)(b+c)$

Substitute this back into the expression for the LHS:

LHS $= 3(a+b)(a+c)(b+c)$

Rearrange the factors in the order given in the RHS:

LHS $= 3(a+b)(b+c)(c+a)$

This is the Right Hand Side (RHS) of the identity.

LHS $=$ RHS

Hence, the identity $(a + b + c)^3 – a^3 – b^3 – c^3 = 3(a + b ) (b + c) (c + a)$ is proven.