Algebraic Identities

Introduction to Algebraic Identities

Equality vs. Identity

In mathematics, specifically in algebra, we frequently work with statements that express equality between two expressions. These statements are generally called equations.

An equation is a mathematical statement that asserts that two expressions have the same value. Equations often involve one or more variables. The crucial aspect of many equations is that they are only true for specific values of the variable(s) involved, called the solutions or roots of the equation. Some equations might have a limited number of solutions, one solution, no solutions, or infinitely many solutions.

Example of an equation that is true for only some values of the variable:

This equation is only true when the value of $x$ is exactly $7$. If $x$ is any other number, the statement is false. The solution set is $\{7\}$.

Example of an equation that is true for no values of the variable:

If we try to solve this equation by subtracting $x$ from both sides, we get $1 = 2$, which is a false statement. No value of $x$ can make this equation true. The solution set is empty ($\emptyset$).

An identity is a very special type of equality statement. Unlike a typical equation, an algebraic identity is an equality between two algebraic expressions that holds true for all possible values that can be substituted for the variables involved, as long as the expressions are defined for those values.

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Definition of an Algebraic Identity

An algebraic identity is an equality statement between two algebraic expressions that is true for every value of the variable(s) for which both expressions are defined.

In essence, an identity represents a fundamental equivalence between two different forms of an algebraic expression. One side of the identity can often be transformed into the other side using the basic properties of arithmetic (like the commutative, associative, and distributive properties).

Example:

This is an algebraic identity. Let's test it with a few values for $x$:

• If $x = 3$: Left side $= 2(3 + 1) = 2(4) = 8$. Right side $= 2(3) + 2 = 6 + 2 = 8$. Left side = Right side.
• If $x = -1$: Left side $= 2(-1 + 1) = 2(0) = 0$. Right side $= 2(-1) + 2 = -2 + 2 = 0$. Left side = Right side.
• If $x = 0$: Left side $= 2(0 + 1) = 2(1) = 2$. Right side $= 2(0) + 2 = 0 + 2 = 2$. Left side = Right side.

No matter what real number we substitute for $x$, the equality $2(x+1) = 2x+2$ will hold. This identity is simply a statement of the distributive property of multiplication over addition.

Identities are like fundamental rules or transformations that can be applied to rewrite algebraic expressions in different equivalent forms, which is extremely useful in simplifying expressions and solving problems.

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The Importance of Algebraic Identities

Algebraic identities are not just mathematical curiosities; they are powerful tools that simplify calculations and manipulations in algebra and beyond. Their importance stems from the fact that they represent universal truths about algebraic expressions.

Key reasons why algebraic identities are important:

1. Simplifying Algebraic Expressions:

   Identities provide standard formulas for expanding products or factoring expressions. For instance, knowing the identity $(a+b)^2 = a^2 + 2ab + b^2$ allows you to quickly expand $(x+3)^2$ as $x^2 + 6x + 9$ without performing the full multiplication $(x+3)(x+3)$. Similarly, identities like $a^2 - b^2 = (a-b)(a+b)$ are essential for factorising expressions like $x^2 - 25$.

2. Performing Algebraic Operations Efficiently:

   Identities can turn complex multiplications or divisions into simpler steps. Multiplying using identities like $(a+b)(a-b) = a^2 - b^2$ is much faster than using direct distribution.

3. Solving Equations:

   Many techniques for solving polynomial equations rely on factoring the polynomial expression. Algebraic identities are crucial tools for factoring, especially for quadratic and some higher-degree polynomials. For example, recognising that $x^2 - 5x + 6$ is in a form related to $(x-a)(x-b)$ helps in factoring it to $(x-2)(x-3)$, which leads to the roots $x=2, x=3$ for the equation $x^2 - 5x + 6 = 0$.

4. Evaluating Numerical Expressions:

   Identities can simplify calculations involving specific numbers. For example, to calculate $103^2$, you could use the identity $(a+b)^2$: $103^2 = (100+3)^2 = 100^2 + 2(100)(3) + 3^2 = 10000 + 600 + 9 = 10609$. This might be easier than direct multiplication $103 \times 103$. Similarly, $49 \times 51 = (50-1)(50+1) = 50^2 - 1^2 = 2500 - 1 = 2499$.

5. Deriving and Proving Other Mathematical Results:

   Algebraic identities serve as fundamental building blocks in more advanced mathematics. They are frequently used in proofs, derivations of formulas (e.g., the quadratic formula), and in various areas of calculus, linear algebra, and abstract algebra.

A solid understanding and memorization of the standard algebraic identities are essential for success in algebra and subsequent mathematical studies. They provide the tools needed to manipulate expressions fluently and solve a wide range of problems efficiently.

Standard Algebraic Identities and their Derivations

Algebraic identities are equality statements that hold true for all values of the variables involved. They are fundamental tools in algebra, simplifying calculations, aiding in factorisation, and forming the basis for more advanced mathematical concepts. These identities are derived directly from the basic axioms and properties of arithmetic operations (like the commutative, associative, and distributive properties).

Here, we will explore some of the most commonly used or "standard" algebraic identities and show how they are derived through algebraic manipulation. In these identities, $a, b,$ and $c$ represent any real numbers.

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Basic Binomial Identities (Degree 2)

These identities involve the square of binomials and the product of a sum and difference.

Identity 1: Square of a Sum

The identity for the square of the sum of two terms is:

Derivation:

By the definition of squaring a quantity, $(a+b)^2$ means $(a+b)$ multiplied by itself:

$$ (a + b)^2 = (a + b) \times (a + b) $$

Now, we use the distributive property to multiply the two binomials. We multiply each term in the first binomial $(a+b)$ by each term in the second binomial $(a+b)$.

$$ (a + b)(a + b) = a(a + b) + b(a + b) $$

Apply the distributive property again to expand the terms:

$$ = (a \times a) + (a \times b) + (b \times a) + (b \times b) $$ $$ = a^2 + ab + ba + b^2 $$

Using the commutative property of multiplication, we know that $ab$ is the same as $ba$. Substitute $ba$ with $ab$:

$$ = a^2 + ab + ab + b^2 $$

Combine the like terms (the two $ab$ terms):

$$ = a^2 + (ab + ab) + b^2 = a^2 + 2ab + b^2 $$

Thus, we have derived the identity: $(a + b)^2 = a^2 + 2ab + b^2$.

Example: Expand $(x+5)^2$. Using the identity with $a=x$ and $b=5$: $(x+5)^2 = x^2 + 2(x)(5) + 5^2 = x^2 + 10x + 25$.

Identity 2: Square of a Difference

The identity for the square of the difference of two terms is:

Derivation:

By the definition of squaring, $(a-b)^2 = (a-b) \times (a-b)$.

Using the distributive property:

$$ (a - b)(a - b) = a(a - b) - b(a - b) $$

Apply the distributive property to expand the terms:

$$ = (a \times a) + (a \times -b) + (-b \times a) + (-b \times -b) $$ $$ = a^2 - ab - ba + b^2 $$

Using the commutative property of multiplication, $ab = ba$. Substitute $ba$ with $ab$:

$$ = a^2 - ab - ab + b^2 $$

Combine the like terms (the two $-ab$ terms):

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

Thus, we have derived the identity: $(a - b)^2 = a^2 - 2ab + b^2$.

Example: Expand $(y-3)^2$. Using the identity with $a=y$ and $b=3$: $(y-3)^2 = y^2 - 2(y)(3) + 3^2 = y^2 - 6y + 9$.

Identity 3: Difference of Squares

The identity for the product of the sum and difference of two terms is:

Derivation:

Using the distributive property to multiply the two binomials $(a+b)$ and $(a-b)$:

$$ (a + b)(a - b) = a(a - b) + b(a - b) $$

Apply the distributive property to expand the terms:

$$ = (a \times a) + (a \times -b) + (b \times a) + (b \times -b) $$ $$ = a^2 - ab + ba - b^2 $$

Using the commutative property of multiplication, $ab = ba$. Substitute $ba$ with $ab$:

$$ = a^2 - ab + ab - b^2 $$

The terms $-ab$ and $+ab$ are additive inverses (they cancel each other out as their sum is $0$).

$$ = a^2 + (-ab + ab) - b^2 = a^2 + 0 - b^2 = a^2 - b^2 $$

Thus, we have derived the identity: $(a + b)(a - b) = a^2 - b^2$.

Example: Expand $(x+4)(x-4)$. Using the identity with $a=x$ and $b=4$: $(x+4)(x-4) = x^2 - 4^2 = x^2 - 16$. This identity is extremely useful for factorisation and simplification.

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Other Useful Identities

Identity 4: Expansion of $(x+a)(x+b)$

This identity describes the product of two binomials where one term is common ($x$) and the other terms are different constants ($a$ and $b$).

Derivation:

Using the distributive property to multiply the two binomials $(x+a)$ and $(x+b)$:

$$ (x + a)(x + b) = x(x + b) + a(x + b) $$

Apply the distributive property again to expand the terms:

$$ = (x \times x) + (x \times b) + (a \times x) + (a \times b) $$ $$ = x^2 + xb + ax + ab $$

Using the commutative property, $xb = bx$ and $ax = xa$. Write the terms with $x$ together:

$$ = x^2 + bx + ax + ab $$

Factor out the common variable $x$ from the two middle terms $bx$ and $ax$. This is an application of the distributive property in reverse:

$$ = x^2 + (b + a)x + ab $$

Using the commutative property of addition, $b+a = a+b$:

$$ = x^2 + (a + b)x + ab $$

Thus, we have derived the identity: $(x + a)(x + b) = x^2 + (a + b)x + ab$.

Example: Expand $(m+2)(m+3)$. Using the identity with $x=m, a=2,$ and $b=3$: $(m+2)(m+3) = m^2 + (2+3)m + (2 \times 3) = m^2 + 5m + 6$.

Identity 5: Square of a Trinomial

This identity expands the square of a sum of three terms.

Derivation:

By definition, $(a+b+c)^2 = (a+b+c)(a+b+c)$.

Using the distributive property, multiply each term in the first trinomial by each term in the second trinomial:

$$ = a(a+b+c) + b(a+b+c) + c(a+b+c) $$

Apply the distributive property to expand each part:

$$ = (a \times a) + (a \times b) + (a \times c) + (b \times a) + (b \times b) + (b \times c) + (c \times a) + (c \times b) + (c \times c) $$ $$ = a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2 $$

Using the commutative property ($ab=ba, ac=ca, bc=cb$), rearrange and group like terms:

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

Combine the like terms:

$$ = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc $$

It is often written with the pairwise product terms in the order $ab, bc, ca$:

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

Thus, we have derived the identity: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

Example: Expand $(x+y+z)^2$. Using the identity with $a=x, b=y,$ and $c=z$: $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(x)(y) + 2(y)(z) + 2(z)(x) = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$.

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Cubic Identities (Degree 3)

These identities involve the cube of binomials and the sum or difference of cubes.

Identity 6: Cube of a Sum

The identity for the cube of the sum of two terms is:

An alternative form by factoring $3ab$ from the middle two terms is: $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$.

Derivation:

By definition, $(a+b)^3 = (a+b)(a+b)^2$. We already derived the identity for $(a+b)^2$: $(a+b)^2 = a^2 + 2ab + b^2$.

Substitute the expanded form of $(a+b)^2$ into the expression for $(a+b)^3$:

$$ (a + b)^3 = (a + b)(a^2 + 2ab + b^2) $$

Using the distributive property, multiply each term in the first binomial by each term in the trinomial:

$$ = a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2) $$

Apply the distributive property again to expand each part:

$$ = (a \times a^2) + (a \times 2ab) + (a \times b^2) + (b \times a^2) + (b \times 2ab) + (b \times b^2) $$ $$ = a^3 + 2a^2b + ab^2 + ba^2 + 2ab^2 + b^3 $$

Using the commutative property ($ba^2 = a^2b$), rearrange and group like terms:

$$ = a^3 + (2a^2b + a^2b) + (ab^2 + 2ab^2) + b^3 $$

Combine the like terms:

$$ = a^3 + 3a^2b + 3ab^2 + b^3 $$

Thus, we have derived the identity: $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

Example: Expand $(x+2)^3$. Using the identity with $a=x$ and $b=2$: $(x+2)^3 = x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3 = x^3 + 6x^2 + 3(x)(4) + 8 = x^3 + 6x^2 + 12x + 8$.

Identity 7: Cube of a Difference

The identity for the cube of the difference of two terms is:

An alternative form by factoring $-3ab$ from the middle two terms is: $(a-b)^3 = a^3 - b^3 - 3ab(a-b)$.

Derivation: This identity can be derived in a similar way to the cube of a sum, by writing $(a-b)^3 = (a-b)(a-b)^2$ and expanding. Alternatively, we can substitute $-b$ for $b$ in the identity for $(a+b)^3$:

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

Simplify the terms:

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

Thus, we have derived the identity: $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$.

Example: Expand $(y-1)^3$. Using the identity with $a=y$ and $b=1$: $(y-1)^3 = y^3 - 3(y^2)(1) + 3(y)(1^2) - 1^3 = y^3 - 3y^2 + 3y - 1$.

Identity 8: Sum of Cubes

This identity provides a way to factor a sum of two cubes into a binomial and a trinomial.

Derivation: To prove this identity, it is easiest to expand the right-hand side (RHS) and show that it is equal to the left-hand side (LHS).

RHS $= (a + b)(a^2 - ab + b^2)$. Using the distributive property:

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

Apply the distributive property again to expand each part:

$$ = (a \times a^2) + (a \times -ab) + (a \times b^2) + (b \times a^2) + (b \times -ab) + (b \times b^2) $$ $$ = a^3 - a^2b + ab^2 + ba^2 - ab^2 + b^3 $$

Using the commutative property ($ba^2 = a^2b$), rearrange and group like terms:

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

The terms $-a^2b$ and $+a^2b$ are additive inverses (they sum to $0$). The terms $ab^2$ and $-ab^2$ are also additive inverses (they sum to $0$).

$$ = a^3 + 0 + 0 + b^3 = a^3 + b^3 $$

This is equal to the LHS. Thus, the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ is proven.

Example: Factor $x^3 + 8$. We recognise $x^3$ as $x^3$ and $8$ as $2^3$. Using the identity with $a=x$ and $b=2$: $x^3 + 8 = x^3 + 2^3 = (x+2)(x^2 - x(2) + 2^2) = (x+2)(x^2 - 2x + 4)$.

Identity 9: Difference of Cubes

This identity provides a way to factor a difference of two cubes into a binomial and a trinomial.

Derivation: Similar to the sum of cubes, we expand the right-hand side (RHS) and show it equals the left-hand side (LHS).

RHS $= (a - b)(a^2 + ab + b^2)$. Using the distributive property:

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

Apply the distributive property to expand each part:

$$ = (a \times a^2) + (a \times ab) + (a \times b^2) - (b \times a^2) - (b \times ab) - (b \times b^2) $$ $$ = a^3 + a^2b + ab^2 - ba^2 - ab^2 - b^3 $$

Using the commutative property ($ba^2 = a^2b$), rearrange and group like terms:

$$ = a^3 + (a^2b - ba^2) + (ab^2 - ab^2) - b^3 $$

Substitute $ba^2 = a^2b$ and note the additive inverses:

$$ = a^3 + (a^2b - a^2b) + (ab^2 - ab^2) - b^3 $$ $$ = a^3 + 0 + 0 - b^3 = a^3 - b^3 $$

This is equal to the LHS. Thus, the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ is proven.

Example: Factor $y^3 - 1$. We recognise $y^3$ as $y^3$ and $1$ as $1^3$. Using the identity with $a=y$ and $b=1$: $y^3 - 1 = y^3 - 1^3 = (y-1)(y^2 + y(1) + 1^2) = (y-1)(y^2 + y + 1)$.

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More Complex Identities (Example)

Identity 10: Sum of Cubes of Three Terms

This identity relates the sum of cubes of three terms to the product of their sum and a quadratic expression.

Derivation: The derivation of this identity is more involved than the previous ones and requires careful expansion of the right-hand side. We can show that $(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = a(a^2 + b^2 + c^2 - ab - bc - ca) + b(a^2 + b^2 + c^2 - ab - bc - ca) + c(a^2 + b^2 + c^2 - ab - bc - ca)$. Expanding this fully and combining like terms leads to $a^3 + b^3 + c^3 - 3abc$. This is left as an exercise in careful algebraic expansion.

Special Case of Identity 10:

A very useful special case arises from Identity 10:

Proof of the Special Case:

If $a + b + c = 0$, substitute $a+b+c=0$ into Identity 10:

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

Add $3abc$ to both sides of the equation:

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

This shows that if the sum of three terms is zero, the sum of their cubes is equal to three times their product.

Example: Without actual cubing, find the value of $(-12)^3 + (7)^3 + (5)^3$.

Let $a=-12, b=7, c=5$. Check the sum $a+b+c$: $-12 + 7 + 5 = -12 + 12 = 0$.

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

$$ (-12)^3 + (7)^3 + (5)^3 = 3 \times (-12) \times (7) \times (5) $$ $$ = 3 \times (-12) \times (35) $$ $$ = 3 \times (-420) = -1260 $$

The value of $(-12)^3 + (7)^3 + (5)^3$ is $-1260$. Calculating the cubes and adding them would be much more tedious.

Memorizing and understanding the derivation of these standard algebraic identities is fundamental for efficiently manipulating algebraic expressions, factorising polynomials, and solving equations.

Using Identities for Simplification and Operations

Understanding and knowing algebraic identities is essential for efficiently manipulating algebraic expressions. Identities are not just statements of equality; they serve as powerful tools that allow us to transform expressions from one form to an equivalent, often simpler, form. They provide shortcuts for performing operations like multiplication and factorisation and can also be used to make numerical calculations easier.

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Applications of Identities

Here are some key ways algebraic identities are used:

1. Expanding Expressions

Identities allow us to quickly expand expressions involving squared or cubed binomials or trinomials without going through the detailed distributive multiplication process. This saves time and reduces the chance of errors.

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2. Factoring Expressions

Algebraic identities are extremely useful for factorising expressions. If an expression matches the expanded form of a standard identity, we can use the identity in reverse to write the expression in its factored form.

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3. Evaluating Numerical Expressions

Algebraic identities can often simplify calculations involving specific numbers, especially large numbers or decimals, by allowing us to rewrite the calculation in a form that is easier to compute mentally or with fewer steps.

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Proficiency in recognising the forms of algebraic expressions that match standard identities and applying the identities (both for expansion and factorisation) is a vital skill that significantly enhances efficiency in algebraic manipulation and problem-solving.