Chapter 4 Exploring Algebraic Identities (Class 9 - Latest Maths NCERT (Ganita Manjari I) Solutions)

To Find:

Expansion of the given expressions using the algebraic identity:

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Solution:

(i) $(7x + 4y)^2$

Here, $a = 7x$ and $b = 4y$.

$(7x + 4y)^2 = (7x)^2 + 2(7x)(4y) + (4y)^2$

$(7x + 4y)^2 = 49x^2 + 56xy + 16y^2$

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(ii) $\left(\frac{7}{5}x + \frac{3}{2}y\right)^2$

Here, $a = \frac{7}{5}x$ and $b = \frac{3}{2}y$.

$\left(\frac{7}{5}x + \frac{3}{2}y\right)^2 = \left(\frac{7}{5}x\right)^2 + 2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right) + \left(\frac{3}{2}y\right)^2$

$\left(\frac{7}{5}x + \frac{3}{2}y\right)^2 = \frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2$

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(iii) $(2.5p + 1.5q)^2$

Here, $a = 2.5p$ and $b = 1.5q$.

$(2.5p + 1.5q)^2 = (2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2$

$(2.5p + 1.5q)^2 = 6.25p^2 + 7.5pq + 2.25q^2$

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(iv) $\left(\frac{3}{4}s + 8t\right)^2$

Here, $a = \frac{3}{4}s$ and $b = 8t$.

$\left(\frac{3}{4}s + 8t\right)^2 = \left(\frac{3}{4}s\right)^2 + 2\left(\frac{3}{4}s\right)(8t) + (8t)^2$

$\left(\frac{3}{4}s + 8t\right)^2 = \frac{9}{16}s^2 + 12st + 64t^2$

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(v) $\left(x + \frac{1}{2y}\right)^2$

Here, $a = x$ and $b = \frac{1}{2y}$.

$\left(x + \frac{1}{2y}\right)^2 = x^2 + 2(x)\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2$

$\left(x + \frac{1}{2y}\right)^2 = x^2 + \frac{x}{y} + \frac{1}{4y^2}$

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(vi) $\left(\frac{1}{x} + \frac{1}{y}\right)^2$

Here, $a = \frac{1}{x}$ and $b = \frac{1}{y}$.

$\left(\frac{1}{x} + \frac{1}{y}\right)^2 = \left(\frac{1}{x}\right)^2 + 2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2$

$\left(\frac{1}{x} + \frac{1}{y}\right)^2 = \frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}$

Solution:

We use the identity $(a + b)^2 = a^2 + 2ab + b^2$ by splitting the numbers into values that are easy to square (like multiples of 10 or 100).

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(i) $(64)^2$

We can write $64$ as $(60 + 4)$.

$(64)^2 = 3600 + 480 + 16$

$(64)^2 = 4096$

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(ii) $(105)^2$

We can write $105$ as $(100 + 5)$.

$(105)^2 = 10000 + 1000 + 25$

$(105)^2 = 11025$

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(iii) $(205)^2$

We can write $205$ as $(200 + 5)$.

$(200 + 5)^2 = 200^2 + 2(200)(5) + 5^2$

$(205)^2 = 40000 + 2000 + 25$

$(205)^2 = 42025$

To Find:

The completely factored form of the given algebraic expressions using the identity:

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Solution:

(i) $9x^2 + 24xy + 16y^2$

We can rewrite the terms as squares:

Checking the middle term: $2(3x)(4y) = 24xy$.

$9x^2 + 24xy + 16y^2 = (3x)^2 + 2(3x)(4y) + (4y)^2$

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

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(ii) $4s^2 + 20st + 25t^2$

Rewriting as squares:

$4s^2 + 20st + 25t^2 = (2s)^2 + 2(2s)(5t) + (5t)^2$

$= (2s + 5t)^2$

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(iii) $49x^2 + 28xy + 4y^2$

Rewriting as squares:

$49x^2 + 28xy + 4y^2 = (7x)^2 + 2(7x)(2y) + (2y)^2$

$= (7x + 2y)^2$

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(iv) $64p^2 + \frac{32}{3}pq + \frac{4}{9}q^2$

Rewriting as squares:

$(8p)^2 + 2(8p)\left(\frac{2}{3}q\right) + \left(\frac{2}{3}q\right)^2$

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

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(v) $3a^2 + 4ab + \frac{4}{3}b^2$

To make the terms perfect squares, we take $\frac{1}{3}$ as a common factor:

Now, factoring the expression inside the bracket:

$\frac{1}{3}[(3a)^2 + 2(3a)(2b) + (2b)^2]$

$= \frac{1}{3}(3a + 2b)^2$

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(vi) $\frac{9}{5}s^2 + 6sv + 5v^2$

Similarly, we take $\frac{1}{5}$ as a common factor:

Now, factoring the expression inside the bracket:

$\frac{1}{5}[(3s)^2 + 2(3s)(5v) + (5v)^2]$

$= \frac{1}{5}(3s + 5v)^2$

To Find:

The squares of the given numbers using the identity:

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Solution:

(i) $(79)^2$

We can write $79$ as $(80 - 1)$.

$(80 - 1)^2 = 80^2 - 2(80)(1) + 1^2$

$(79)^2 = 6400 - 160 + 1$

$(79)^2 = 6241$

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(ii) $(193)^2$

We can write $193$ as $(200 - 7)$.

$(200 - 7)^2 = 200^2 - 2(200)(7) + 7^2$

$(193)^2 = 40000 - 2800 + 49$

$(193)^2 = 37249$

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(iii) $(299)^2$

We can write $299$ as $(300 - 1)$.

$(300 - 1)^2 = 300^2 - 2(300)(1) + 1^2$

$(299)^2 = 90000 - 600 + 1$

$(299)^2 = 89401$

To Find: Squares of the given numbers using algebraic identities.

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Solution:

We use the following identities to make the calculations easier:

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(i) $117^2$: We can write $117$ as $(100 + 17)$. Using identity (i):

$117^2 = (100 + 17)^2 = 100^2 + 2(100)(17) + 17^2$

$117^2 = 10000 + 3400 + 289 = \mathbf{13689}$

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(ii) $78^2$: We can write $78$ as $(80 - 2)$. Using identity (ii):

$78^2 = (80 - 2)^2 = 80^2 - 2(80)(2) + 2^2$

$78^2 = 6400 - 320 + 4 = \mathbf{6084}$

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(iii) $198^2$: We can write $198$ as $(200 - 2)$. Using identity (ii):

$198^2 = (200 - 2)^2 = 200^2 - 2(200)(2) + 2^2$

$198^2 = 40000 - 800 + 4 = \mathbf{39204}$

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(iv) $214^2$: We can write $214$ as $(200 + 14)$. Using identity (i):

$214^2 = (200 + 14)^2 = 200^2 + 2(200)(14) + 14^2$

$214^2 = 40000 + 5600 + 196 = \mathbf{45796}$

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(v) $1104^2$: We can write $1104$ as $(1100 + 4)$. Using identity (i):

$1104^2 = (1100 + 4)^2 = 1100^2 + 2(1100)(4) + 4^2$

$1104^2 = 1210000 + 8800 + 16 = \mathbf{1218816}$

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(vi) $1120^2$: We can write $1120$ as $(1100 + 20)$. Using identity (i):

$1120^2 = (1100 + 20)^2 = 1100^2 + 2(1100)(20) + 20^2$

$1120^2 = 1210000 + 44000 + 400 = \mathbf{1254400}$

To factorize: The given algebraic expressions using perfect square identities.

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(i) $16y^2 – 24y + 9$

This is in the form of $a^2 - 2ab + b^2$ where $a = 4y$ and $b = 3$.

$(4y)^2 - 2(4y)(3) + 3^2 = \mathbf{(4y - 3)^2}$

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(ii) $\frac{9}{4}s^2 + 6st + 4t^2$

This is in the form of $a^2 + 2ab + b^2$ where $a = \frac{3}{2}s$ and $b = 2t$.

$(\frac{3}{2}s)^2 + 2(\frac{3}{2}s)(2t) + (2t)^2 = \mathbf{(\frac{3}{2}s + 2t)^2}$

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(iii) $\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2$

Rearranging the terms to group squares and mixed products:

$= (\frac{m}{3})^2 + (\frac{k}{2})^2 + (3n)^2 + 2(\frac{m}{3})(\frac{k}{2}) + 2(\frac{k}{2})(3n) + 2(\frac{m}{3})(3n)$

This matches the identity $(a+b+c)^2 = a^2+b^2+c^2 $$ +2ab+2bc+2ca$.

$= \mathbf{(\frac{m}{3} + \frac{k}{2} + 3n)^2}$

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(iv) $\frac{p^2}{16} - 2 + \frac{16}{p^2}$

This is $a^2 - 2ab + b^2$ where $a = \frac{p}{4}$ and $b = \frac{4}{p}$.

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

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(v) $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc$

Observing the negative signs, terms involving $b$ produce negative products with $a$ and $c$. Thus, $2b$ is negative.

$(3a)^2 + (-2b)^2 + c^2 + 2(3a)(-2b) + 2(-2b)(c) + 2(3a)(c)$

$= \mathbf{(3a - 2b + c)^2}$

Solution:

(i) $(p + 3q + 7r)^2$

Here $a=p, b=3q, c=7r$.

$= p^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(p)(7r)$

$= \mathbf{p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr}$

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(ii) $(3x – 2y + 4z)^2$

Here $a=3x, b=-2y, c=4z$.

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

$= \mathbf{9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz}$

To Check: If the given equation is an identity.

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Solution:

Let us expand the Left Hand Side (LHS) of the equation:

Adding (1), (2), and (3):

$\text{LHS} = (a^2+b^2+c^2+2ab-2bc-2ca) $$ + (a^2+b^2+c^2-2ab-2bc+2ca) $$ + (a^2+b^2+c^2-2ab+2bc-2ca)$

$\text{LHS} = 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ca$

However, the Right Hand Side (RHS) is given as $2a^2 + 2b^2 + 2c^2$.

Comparing LHS and RHS:

Since the LHS is not equal to the RHS for all values of $a, b,$ and $c$, this is not an identity.

To Find:

Complete the factorization of the given quadratic expressions by filling in the blanks.

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Solution:

(i) $s^2 - 11s + 24$:

We need to find two numbers whose product is $24$ and whose sum is $-11$. These numbers are $-3$ and $-8$.

$s^2 - 3s - 8s + 24 = s(s - 3) - 8(s - 3)$

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(ii) $(\_\_\_\_\_\_\_\_) (x + 1) = (3x^2 – 4x – 7)$:

We factorize $3x^2 - 4x - 7$ by splitting the middle term (product = $-21$, sum = $-4$):

$3x^2 + 3x - 7x - 7 = 3x(x + 1) - 7(x + 1)$

The blank is $(3x - 7)$.

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(iii) $10x^2 – 11x – 6 = (2x – \_\_\_) (\_\_\_ + 2)$:

Factorizing $10x^2 - 11x - 6$ (product = $-60$, sum = $-11$):

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

The blanks are $3$ and $5x$.

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(iv) $6x^2 + 7x + 2 = (\_\_\_\_\_\_\_\_\_\_\_\_) (\_\_\_\_\_\_\_\_\_\_\_)$:

Factorizing $6x^2 + 7x + 2$ (product = $12$, sum = $7$):

$6x^2 + 4x + 3x + 2 = 2x(3x + 2) + 1(3x + 2)$

Solution:

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(i) $(41)^2$: Use $(a + b)^2 = a^2 + 2ab + b^2$

$(40 + 1)^2 = 40^2 + 2(40)(1) + 1^2 = 1600 + 80 + 1 = \mathbf{1681}$

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(ii) $(27)^2$: Use $(a - b)^2 = a^2 - 2ab + b^2$

$(30 - 3)^2 = 30^2 - 2(30)(3) + 3^2 = 900 - 180 + 9 = \mathbf{729}$

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(iii) $(23 \times 17)$: Use $(a + b)(a - b) = a^2 - b^2$

$(20 + 3)(20 - 3) = 20^2 - 3^2 = 400 - 9 = \mathbf{391}$

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(iv) $(135)^2$: Use $(a + b)^2$

$(130 + 5)^2 = 130^2 + 2(130)(5) + 5^2 = 16900 + 1300 + 25 $$ = \mathbf{18225}$

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(v) $(97)^2$: Use $(a - b)^2$

$(100 - 3)^2 = 100^2 - 2(100)(3) + 3^2 = 10000 - 600 + 9 = \mathbf{9409}$

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(vi) $(18 \times 29)$: Use $(x + a)(x + b) = x^2 + (a+b)x + ab$

Taking $x = 20$: $(20 - 2)(20 + 9) = 20^2 + (-2 + 9)20 + (-2)(9) $$ = 400 + 140 - 18 = \mathbf{522}$

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(vii) $(34 \times 43)$: Use $(x + a)(x + b)$

Taking $x = 40$: $(40 - 6)(40 + 3) = 40^2 + (-6 + 3)40 + (-6)(3) $$ = 1600 - 120 - 18 = \mathbf{1462}$

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(viii) $(205)^2$: Use $(a + b)^2$

$(200 + 5)^2 = 200^2 + 2(200)(5) + 5^2 = 40000 + 2000 + 25 $$ = \mathbf{42025}$

Solution:

(i) $9a^2 + b^2 + 4c^2 – 6ab + 12ac – 4bc$:

Using $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$. Looking at the signs, terms with '$b$' are negative.

$= (3a)^2 + (-b)^2 + (2c)^2 + 2(3a)(-b) + 2(-b)(2c) + 2(3a)(2c)$

$= (3a - b + 2c)^2$

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(ii) $16s^2 + 25t^2 – 40st$:

$= (4s)^2 + (5t)^2 - 2(4s)(5t)$

$= (4s - 5t)^2$

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(iii) $r^2 – r – 42$:

Factors of $-42$ that sum to $-1$ are $-7$ and $6$.

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

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

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(iv) $49g^2 + 14gh + h^2$:

$= (7g)^2 + 2(7g)(h) + (h)^2$

$= (7g + h)^2$

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(v) $64u^2 + 121v^2 + 4w^2 – 176uv – 32uw + 44vw$:

Using $(x + y + z)^2$. Signs indicate '$u$' is negative while '$v$' and '$w$' are positive (as $vw$ is positive but $uv$ and $uw$ are negative).

$= (-8u)^2 + (11v)^2 + (2w)^2 + 2(-8u)(11v) + 2(11v)(2w) $$ + 2(-8u)(2w)$

$= (-8u + 11v + 2w)^2$ or $(8u - 11v - 2w)^2$

To Find:

We need to simplify each of the given rational expressions by factorising the numerators and denominators and cancelling out common factors.

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(i) $\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}$

Solution:

First, we factorise the numerator by taking out the common factor $3$:

$3p^2 - 3pq - 18q^2 = 3(p^2 - pq - 6q^2)$

Splitting the middle term of the quadratic expression $p^2 - pq - 6q^2$:

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

So, the numerator becomes $3(p - 3q)(p + 2q)$.

Now, we factorise the denominator $p^2 + 3pq - 10q^2$ by splitting the middle term:

$p^2 + 5pq - 2pq - 10q^2 = p(p + 5q) - 2q(p + 5q) $$ = (p - 2q)(p + 5q)$

Substituting these back into the original expression:

$\frac{3(p - 3q)(p + 2q)}{(p - 2q)(p + 5q)}$

Since there are no common factors to cancel out, the simplified form is:

$\mathbf{\frac{3(p - 3q)(p + 2q)}{(p - 2q)(p + 5q)}}$

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(ii) $\frac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}$

Solution:

The numerator $n^3 - 3n^2m + 3nm^2 - m^3$ follows the algebraic identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$. Thus:

The denominator can be factorised by taking out $5$ as a common factor:

$5(m^2 - 2mn + n^2)$

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

Note that $(n - m)^3 = [-(m - n)]^3 = -(m - n)^3$. Substituting (i) and (ii):

$\frac{-(m - n)^3}{5(m - n)^2} = \frac{-(m - n)}{5} = \frac{n - m}{5}$

Final Answer: $\mathbf{\frac{n - m}{5}}$

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(iii) $\frac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}$

Solution:

We use the identity $a^3 + b^3 + c^3 - 3abc $$ = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$. Here $a = w, b = -v, c = x$.

Numerator: $(w - v + x)(w^2 + (-v)^2 + x^2 - w(-v) - (-v)x - wx)$

$= (w - v + x)(w^2 + v^2 + x^2 + wv + vx - wx)$

Denominator follows the identity $(a + b + c)^2 $$ = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$. Here $a = w, b = -v, c = x$.

Denominator: $(w - v + x)^2$

Substituting these:

$\frac{(w - v + x)(w^2 + v^2 + x^2 + wv + vx - wx)}{(w - v + x)^2}$

$\frac{\cancel{(w - v + x)}(w^2 + v^2 + x^2 + wv + vx - wx)}{\cancel{(w - v + x)}(w - v + x)}$

Final Answer: $\mathbf{\frac{w^2 + v^2 + x^2 + wv + vx - wx}{w - v + x}}$

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(iv) $\frac{4y^2 - 20yz + 25z^2}{(25z^2 - 4y^2)}$

Solution:

Numerator is a perfect square: $(2y)^2 - 2(2y)(5z) + (5z)^2 $$ = (2y - 5z)^2$.

Since $(a - b)^2 = (b - a)^2$, we can write this as $(5z - 2y)^2$.

Denominator is a difference of squares: $(5z)^2 - (2y)^2 $$ = (5z - 2y)(5z + 2y)$.

Expression: $\frac{(5z - 2y)^2}{(5z - 2y)(5z + 2y)}$

Cancelling the common factor $(5z - 2y)$:

$\frac{\cancel{(5z - 2y)}(5z - 2y)}{\cancel{(5z - 2y)}(5z + 2y)}$

Final Answer: $\mathbf{\frac{5z - 2y}{5z + 2y}}$

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(v) $\frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}$

Solution:

Factorise each quadratic expression separately:

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

2. $x^2 - 7x + 12 = (x - 3)(x - 4)$

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

4. $x^2 - 9 = (x - 3)(x + 3)$

Substituting into the rational expression:

$\frac{(x + 3)(x - 2)(x - 3)(x - 4)}{(x - 2)(x - 4)(x - 3)(x + 3)}$

Cancelling all common factors in numerator and denominator:

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

Final Answer: $\mathbf{1}$

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(vi) $\frac{p^4 - 16}{p^2 - 4p + 4}$

Solution:

Numerator: $(p^2)^2 - (4)^2 = (p^2 - 4)(p^2 + 4)$

Further factorising $(p^2 - 4)$: $(p - 2)(p + 2)(p^2 + 4)$.

Denominator: $(p - 2)^2$ (Perfect square identity).

Expression: $\frac{(p - 2)(p + 2)(p^2 + 4)}{(p - 2)(p - 2)}$

Cancelling $(p - 2)$ from numerator and denominator:

$\mathbf{\frac{(p + 2)(p^2 + 4)}{p - 2}}$

To Find:

The products of the given algebraic expressions using suitable identities.

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Solution:

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

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, where $a = -3x$ and $b = 4$:

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

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(ii) $(2s + 7) (2s – 7)$

Using the identity $(a + b)(a - b) = a^2 - b^2$, where $a = 2s$ and $b = 7$:

$(2s + 7) (2s – 7) = (2s)^2 - (7)^2$

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(iii) $(p^2 + \frac{1}{2}) (p^2 – \frac{1}{2})$

Using the identity $(a + b)(a - b) = a^2 - b^2$, where $a = p^2$ and $b = \frac{1}{2}$:

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

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(iv) $(2n + 7) (2n – 7)$

Using $(a + b)(a - b) = a^2 - b^2$, where $a = 2n$ and $b = 7$:

$(2n + 7) (2n – 7) = (2n)^2 - (7)^2$

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(v) $(s – 2t) (s^2 + 2st + 4t^2)$

Using the identity $(a - b)(a^2 + ab + b^2) = a^3 - b^3$, where $a = s$ and $b = 2t$:

$(s – 2t) (s^2 + 2st + (2t)^2) = s^3 - (2t)^3$

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(vi) $\left( \frac{1}{2r} - 4r \right)^2$

Using $(a - b)^2 = a^2 - 2ab + b^2$, where $a = \frac{1}{2r}$ and $b = 4r$:

$\left( \frac{1}{2r} - 4r \right)^2 = \left( \frac{1}{2r} \right)^2 - 2\left( \frac{1}{2r} \right)(4r) + (4r)^2$

$\left( \frac{1}{2r} - 4r \right)^2 = \frac{1}{4r^2} - 4 + 16r^2$

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(vii) $(-3m + 4k – l)^2$

Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$, where $a = -3m, b = 4k, c = -l$:

$(-3m)^2 + (4k)^2 + (-l)^2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-3m)(-l)$

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(viii) $(x - \frac{1}{3}y)^3$

Using $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, where $a = x$ and $b = \frac{1}{3}y$:

$(x - \frac{1}{3}y)^3 = (x)^3 - 3(x)^2(\frac{1}{3}y) + 3(x)(\frac{1}{3}y)^2 - (\frac{1}{3}y)^3$

$= x^3 - 3x^2(\frac{1}{3}y) + 3x(\frac{1}{9}y^2) - \frac{1}{27}y^3$

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(ix) $(\frac{7}{2}k - \frac{2}{3}m)^3$

Using $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, where $a = \frac{7}{2}k$ and $b = \frac{2}{3}m$:

$(\frac{7}{2}k)^3 - 3(\frac{7}{2}k)^2(\frac{2}{3}m) + 3(\frac{7}{2}k)(\frac{2}{3}m)^2 - (\frac{2}{3}m)^3$

$= \frac{343}{8}k^3 - 3(\frac{49}{4}k^2)(\frac{2}{3}m) + 3(\frac{7}{2}k)(\frac{4}{9}m^2) - \frac{8}{27}m^3$

To Find:

The values of the given expressions using algebraic identities.

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Solution:

(i) $17 \times 21$

We can rewrite the numbers as $(19 - 2)$ and $(19 + 2)$.

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

$17 \times 21 = 361 - 4$

Result: $357$

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(ii) $104 \times 96$

We can rewrite the numbers as $(100 + 4)$ and $(100 - 4)$.

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

$104 \times 96 = 100^2 - 4^2$

$104 \times 96 = 10000 - 16$

Result: $9984$

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(iii) $24 \times 16$

We can rewrite the numbers as $(20 + 4)$ and $(20 - 4)$.

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

$24 \times 16 = 20^2 - 4^2$

$24 \times 16 = 400 - 16$

Result: $384$

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(iv) $147^3$

We can rewrite $147$ as $(150 - 3)$.

Using the identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

$147^3 = (150 - 3)^3$

$147^3 = 150^3 - 3(150^2)(3) + 3(150)(3^2) - 3^3$

$147^3 = 3375000 - 9(22500) + 450(9) - 27$

$147^3 = 3375000 - 202500 + 4050 - 27$

Result: $3176523$

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(v) $199^3$

We can rewrite $199$ as $(200 - 1)$.

Using the identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

$199^3 = (200 - 1)^3$

$199^3 = 200^3 - 3(200^2)(1) + 3(200)(1^2) - 1^3$

$199^3 = 8000000 - 120000 + 600 - 1$

Result: $7880599$

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(vi) $127^3$

We can rewrite $127$ as $(130 - 3)$.

Using the identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

$127^3 = (130 - 3)^3$

$127^3 = 130^3 - 3(130^2)(3) + 3(130)(3^2) - 3^3$

$127^3 = 2197000 - 152100 + 3510 - 27$

Result: $2048383$

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(vii) $(-107)^3$

We can write this as $-[(100 + 7)^3]$.

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

$(100 + 7)^3 = 100^3 + 3(100^2)(7) + 3(100)(7^2) + 7^3$

$(100 + 7)^3 = 1000000 + 210000 + 14700 + 343$

$(100 + 7)^3 = 1225043$

Result: $-1225043$

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(viii) $(-299)^3$

We can write this as $-[(300 - 1)^3]$.

Using the identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

$(300 - 1)^3 = 300^3 - 3(300^2)(1) + 3(300)(1^2) - 1^3$

$(300 - 1)^3 = 27000000 - 270000 + 900 - 1$

$(300 - 1)^3 = 26730899$

Result: $-26730899$

To Factorise:

Algebraic expressions using standard identities.

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(i) $4y^2 + 1 + \frac{1}{16y^2}$

This expression is in the form $a^2 + 2ab + b^2$.

Using $(a + b)^2 = a^2 + 2ab + b^2$, where $a = 2y$ and $b = \frac{1}{4y}$:

Result: $\mathbf{\left(2y + \frac{1}{4y}\right)^2}$

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(ii) $9m^2 - \frac{1}{25n^2}$

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

$= (3m)^2 - \left(\frac{1}{5n}\right)^2$

Result: $\mathbf{\left(3m - \frac{1}{5n}\right)\left(3m + \frac{1}{5n}\right)}$

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(iii) $27b^3 - \frac{1}{64b^3}$

Using the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, where $a = 3b$ and $b = \frac{1}{4b}$:

$= (3b)^3 - \left(\frac{1}{4b}\right)^3$

$= \left(3b - \frac{1}{4b}\right) \left[ (3b)^2 + (3b)\left(\frac{1}{4b}\right) + \left(\frac{1}{4b}\right)^2 \right]$

Result: $\mathbf{\left(3b - \frac{1}{4b}\right) \left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)}$

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(iv) $x^2 + \frac{5}{6}x + \frac{1}{6}$

We split the middle term using two numbers whose sum is $\frac{5}{6}$ and product is $\frac{1}{6}$. The numbers are $\frac{1}{2}$ and $\frac{1}{3}$.

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

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

Result: $\mathbf{\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)}$

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(v) $27u^3 - \frac{1}{125} - \frac{27u^2}{5} + \frac{9u}{25}$

Using identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, where $a = 3u$ and $b = \frac{1}{5}$:

Result: $\mathbf{\left(3u - \frac{1}{5}\right)^3}$

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(vi) $64y^3 + \frac{1}{125}z^3$

Using identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, where $a = 4y$ and $b = \frac{z}{5}$:

$= (4y)^3 + \left(\frac{z}{5}\right)^3$

Result: $\mathbf{\left(4y + \frac{z}{5}\right) \left(16y^2 - \frac{4yz}{5} + \frac{z^2}{25}\right)}$

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(vii) $p^3 + 27q^3 + r^3 - 9pqr$

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

$= (p)^3 + (3q)^3 + (r)^3 - 3(p)(3q)(r)$

Result: $\mathbf{(p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - pr)}$

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(viii) $9m^2 - 12m + 4$

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

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

Result: $\mathbf{(3m - 2)^2}$

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(ix) $9x^3 - \frac{8}{3}y^3 + \frac{z^3}{3} + 6xyz$

Taking $\frac{1}{3}$ as a common factor:

$= \frac{1}{3}(27x^3 - 8y^3 + z^3 + 18xyz)$

$= \frac{1}{3}[(3x)^3 + (-2y)^3 + (z)^3 - 3(3x)(-2y)(z)]$

Result: $\mathbf{\frac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3xz)}$

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(x) $4x^2 + 9y^2 + 36z^2 + 12xy + 36yz + 24xz$

Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$:

$= (2x)^2 + (3y)^2 + (6z)^2 + 2(2x)(3y) + 2(3y)(6z) + 2(2x)(6z)$

Result: $\mathbf{(2x + 3y + 6z)^2}$

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(xi) $27u^3 - \frac{1}{216} - \frac{9u^2}{2} + \frac{u}{4}$

Using $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, where $a = 3u$ and $b = \frac{1}{6}$:

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

Result: $\mathbf{\left(3u - \frac{1}{6}\right)^3}$

To Find:

We need to simplify three rational algebraic expressions by factorising the numerators and denominators and cancelling out common factors.

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(i) $\frac{4x^2 + 4x + 1}{4x^2 - 1}$

Solution:

First, let us factorise the numerator $4x^2 + 4x + 1$. We observe that it is a perfect square trinomial:

Next, we factorise the denominator $4x^2 - 1$. This is in the form of difference of squares, $a^2 - b^2 = (a - b)(a + b)$:

Substituting the values from equations (i) and (ii) back into the given expression:

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

By cancelling the common factor $(2x + 1)$ from the numerator and denominator, we get:

Final Answer: $\mathbf{\frac{2x + 1}{2x - 1}}$

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(ii) $\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}$

Solution:

First, we simplify the numerator by taking out common factors:

$9(3a^3 - 24b^3) = 9 \times 3(a^3 - 8b^3) = 27(a^3 - (2b)^3)$

Using the identity for difference of cubes, $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$:

Now, we factorise the denominator $9a^2 - 36b^2$ by taking out the common factor $9$:

$9(a^2 - 4b^2) = 9(a^2 - (2b)^2)$

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

Substituting (i) and (ii) back into the expression:

$\frac{27(a - 2b)(a^2 + 2ab + 4b^2)}{9(a - 2b)(a + 2b)}$

Cancelling $9$ with $27$ (which gives $3$) and the common factor $(a - 2b)$:

Final Answer: $\mathbf{\frac{3(a^2 + 2ab + 4b^2)}{a + 2b}}$

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(iii) $\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}$

Solution:

We factorise the numerator using the identity for sum of cubes, $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$:

Now, we factorise the denominator $s^2 - 2st - 35t^2$ by splitting the middle term. We need two numbers whose sum is $-2$ and product is $-35$. These numbers are $-7$ and $5$:

$s^2 - 7st + 5st - 35t^2$

$s(s - 7t) + 5t(s - 7t)$

Substituting (i) and (ii) back into the expression:

$\frac{(s + 5t)(s^2 - 5st + 25t^2)}{(s + 5t)(s - 7t)}$

Cancelling the common factor $(s + 5t)$:

Final Answer: $\mathbf{\frac{s^2 - 5st + 25t^2}{s - 7t}}$

To Find: Dimensions (Length and Breadth) of the rectangles from the given area.

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(i) Area = $25a^2 – 30ab + 9b^2$

We observe that this is a perfect square trinomial:

$25a^2 – 30ab + 9b^2 = (5a)^2 - 2(5a)(3b) + (3b)^2$

Possible expressions are:

Length = $\mathbf{5a - 3b}$ units

Breadth = $\mathbf{5a - 3b}$ units

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(ii) Area = $36s^2 – 49t^2$

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

$36s^2 – 49t^2 = (6s)^2 - (7t)^2$

Possible expressions are:

Length = $\mathbf{6s + 7t}$ units

Breadth = $\mathbf{6s - 7t}$ units

To Find: Dimensions (Length, Breadth, and Height) of the cuboids from the given volume.

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(i) Volume = $6a^2 – 24b^2$

Taking $6$ as a common factor:

$6(a^2 - 4b^2) = 6[(a)^2 - (2b)^2]$

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

Possible expressions for dimensions are:

$\mathbf{6, (a - 2b), \text{ and } (a + 2b)}$ units

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(ii) Volume = $3ps^2 – 15ps + 12p$

Taking $3p$ as a common factor:

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

Factorizing the quadratic part $s^2 - 5s + 4$ by splitting the middle term:

$s^2 - 4s - s + 4 = s(s - 4) - 1(s - 4) = (s - 4)(s - 1)$

Possible expressions for dimensions are:

$\mathbf{3p, (s - 4), \text{ and } (s - 1)}$ units

Given:

Side of the square playground = $40\text{ m}$

Width of the path around the playground = $s\text{ m}$

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To Find:

An expression for the area of the path in terms of $s$.

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Solution:

The path is created around the playground. Let the side of the inner square be $x$ and the side of the outer square (including the path) be $y$.

Since the path is added to both sides of the playground, the outer side $y$ is:

$y = 40 + s + s = 40 + 2s$

The area of the path is the difference between the area of the outer square and the area of the inner square.

$\text{Area of path} = (\text{Outer side})^2 - (\text{Inner side})^2$

$\text{Area of path} = (40 + 2s)^2 - (40)^2$

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

$\text{Area of path} = (40^2 + 2 \times 40 \times 2s + (2s)^2) - 1600$

$\text{Area of path} = (1600 + 160s + 4s^2) - 1600$

By factoring out $4s$, we can also write it as:

$\text{Area of path} = 4s(40 + s)\text{ square metres}$

To Find:

A number $x$ such that the sum of the number and its reciprocal is $\frac{10}{3}$.

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Solution:

Let the required number be $x$. Its reciprocal is $\frac{1}{x}$.

According to the question:

Multiplying the entire equation by $3x$ to remove the denominators:

$3x(x) + 3x\left(\frac{1}{x}\right) = 3x\left(\frac{10}{3}\right)$

$3x^2 + 3 = 10x$

$3x^2 - 10x + 3 = 0$

We solve this quadratic equation by splitting the middle term. We need two numbers whose product is $3 \times 3 = 9$ and sum is $-10$. These numbers are $-9$ and $-1$.

$3x^2 - 9x - x + 3 = 0$

$3x(x - 3) - 1(x - 3) = 0$

$(3x - 1)(x - 3) = 0$

This gives us two possible values for $x$:

Therefore, the number is either 3 or $\frac{1}{3}$.

Given:

Area of rectangular pool = $2x^2 + 7x + 3\text{ square hastas}$

Width of the pool = $2x + 1\text{ hastas}$

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To Find:

The length of the pool.

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Solution:

In Indian perspective, Hasta is an ancient unit of length (approximately the length of the forearm). The area of a rectangle is the product of its length and width.

$\text{Length} = \frac{\text{Area}}{\text{Width}}$

$\text{Length} = \frac{2x^2 + 7x + 3}{2x + 1}$

To simplify, let us factorize the quadratic expression $2x^2 + 7x + 3$:

$2x^2 + 7x + 3 = 2x^2 + 6x + x + 3$

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

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

Substituting this back into the length expression:

$\text{Length} = \frac{\cancel{(2x + 1)}(x + 3)}{\cancel{2x + 1}}$

Therefore, the length of the pool is $(x + 3)$ hastas.

Given:

A polynomial $f(x) = px^2 + 5x + r$.

The factors are $(x - 2)$ and $(x - \frac{1}{2})$.

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To Prove:

$p = r$.

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Proof:

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

Since $(x - 2)$ is a factor:

$f(2) = p(2)^2 + 5(2) + r = 0$

Since $(x - \frac{1}{2})$ is a factor:

$f\left(\frac{1}{2}\right) = p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r = 0$

$\frac{p}{4} + \frac{5}{2} + r = 0$

Multiplying by $4$ to clear the fractions:

From equations (i) and (ii), since both are equal to zero, we can equate them:

Subtracting $10$ from both sides:

$4p + r = p + 4r$

$4p - p = 4r - r$

$3p = 3r$

Given:

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To Prove:

$a^3 + b^3 + c^3 – 3abc = -25$

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Proof:

We know the algebraic identity:

First, we need to find the value of $a^2 + b^2 + c^2$. Using the identity for the square of a trinomial:

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

Substituting values from (i) and (ii):

$(5)^2 = a^2 + b^2 + c^2 + 2(10)$

$25 = a^2 + b^2 + c^2 + 20$

Now, substituting the values of (i), (ii), and (iii) into the identity $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca))$:

$a^3 + b^3 + c^3 - 3abc = (5) \times (5 - 10)$

$a^3 + b^3 + c^3 - 3abc = 5 \times (-5)$

Given:

The expression $n^3 - n$, where $n$ is any natural number ($1, 2, 3, \dots$).

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To Prove:

The expression $n^3 - n$ is always divisible by $6$.

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Proof:

First, we factorise the given algebraic expression by taking $n$ as a common factor:

$n^3 - n = n(n^2 - 1)$

Using the identity $a^2 - b^2 = (a - b)(a + b)$, we can further factorise $(n^2 - 1)$ as $(n - 1)(n + 1)$:

From equation (i), we can see that $n^3 - n$ is the product of three consecutive natural numbers.

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Reasons for Divisibility:

For any number to be divisible by $6$, it must be divisible by both its prime factors, $2$ and $3$.

1. Divisibility by 2:

In any set of two consecutive numbers, one number is always even (divisible by $2$). Since the expression contains three consecutive numbers, at least one of them must be even. Therefore, the product is always divisible by $2$.

2. Divisibility by 3:

In any set of three consecutive numbers, exactly one number must be a multiple of 3. Thus, their product $(n - 1)n(n + 1)$ is always divisible by $3$.

3. Divisibility by 6:

Since the expression is always divisible by $2$ and $3$, and we know that $2$ and $3$ are co-prime numbers, their product must be divisible by $2 \times 3 = 6$.

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Conclusion:

As the factorised form $(n - 1)n(n + 1)$ represents the product of three consecutive integers, it satisfies the conditions for divisibility by $2$ and $3$ simultaneously. Hence, $n^3 - n$ is always divisible by $6$ for all natural numbers $n$.

Solution (i):

Given: $x + y = -4$

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

Let $a = x$, $b = y$, and $c = 4$. Since $x + y + 4 = 0$:

$x^3 + y^3 + 4^3 = 3(x)(y)(4)$

$x^3 + y^3 + 64 = 12xy$

The value of the expression is 0.

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Solution (ii):

Given: $x = 2y + 6$

Let $a = x$, $b = -2y$, and $c = -6$. Let's check their sum:

$a + b + c = x + (-2y) + (-6) = x - 2y - 6$

From (ii), we know the sum is 0. Thus:

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

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

$x^3 - 8y^3 - 216 = 36xy$

The value of the expression is 0.