Algebraic Identities

The algebraic equations which are true for all values of variables in them are called algebraic identities. They are also used for the factorization of polynomials. This is how they find utility in the computation of algebraic expressions. You have already learned about a few of them in the junior grades. In this article, we will recall them and introduce you to some more standard algebraic identities along with examples.

Algebraic Identities

Standard Algebraic Identities List

All the standard Algebraic Identities are derived from the Binomial Theorem, which is given as:

\( \mathbf{(a+b)^{n} =\; ^{n}C_{0}.a^{n}.b^{0} +^{n} C_{1} . a^{n-1} . b^{1} + …….. + ^{n}C_{n-1}.a^{1}.b^{n-1} + ^{n}C_{n}.a^{0}.b^{n}}\)

Some Standard Algebraic Identities list are given below:

Identity I: (a + b)2 = a2 + 2ab + b2

Identity II: (a – b)2 = a2 – 2ab + b2

Identity III: a2 – b2= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab

Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)

Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)

Identity VIII: a3 + b3 + c– 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

 

Example 1: Find the product of (x + 1)(x + 1) using standard algebraic identities.

Solution: (x + 1)(x + 1) can be written as (x + 1)2. Thus, it is of the form Identity I where a = x and b = 1. So we have,

(x + 1)2 = (x)2 + 2(x)(1) + (1)2 = x2 + 2x + 1

 

Example 2: Factorise (x4 – 1) using standard algebraic identities.

Solution: (x4 – 1) is of the form Identity III where a = x2 and b = 1. So we have,

(x4 – 1) = ((x2)2– 12) = (x2 + 1)(x2 – 1)

The factor (x2 – 1) can be further factorised using the same Identity III where a = x and b = 1. So,

(x4 – 1) = (x2 + 1)((x)2 –(1)2) = (x2 + 1)(x + 1)(x – 1)

 

Eample 3: Factorise 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx using standard algebraic identities.

Solution: 16x2 + 4y2 + 9z2– 16xy + 12yz – 24zx is of the form Identity V. So we have,

16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx = (4x)2 + (-2y)2 + (-3z)2 + 2(4x)(-2y) + 2(-2y)(-3z) + 2(-3z)(4x)= (4x – 2y – 3z)2 = (4x – 2y – 3z)(4x – 2y – 3z)

 

Example 4: Expand (3x – 4y)3 using standard algebraic identities.

Solution: (3x– 4y)3 is of the form Identity VII where a = 3x and b = 4y. So we have,

(3x – 4y)3 = (3x)3 – (4y)3– 3(3x)(4y)(3x – 4y) = 27x3 – 64y3 – 108x2y + 144xy2

 

Example 5: Factorize (x3 + 8y3 + 27z3 – 18xyz) using standard algebraic identities.

Solution: (x3 + 8y3 + 27z3 – 18xyz)is of the form Identity VIII where a = x, b = 2y and c = 3z. So we have,

(x3 + 8y3 + 27z3 – 18xyz) = (x)3 + (2y)3 + (3z)3 – 3(x)(2y)(3z)= (x + 2y + 3z)(x2 + 4y2 + 9z2 – 2xy – 6yz – 3zx)

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Frequently Asked Questions on Algebraic Identities

What are three algebraic identities in Maths?

The three algebraic identities in Maths are:

Identity 1: (a+b)^2 = a^2 + b^2 + 2ab

Identity 2: (a-b)^2 = a^2 + b^2 – 2ab

Identity 3: a^2 – b^2 = (a+b) (a-b)

What is the difference between an algebraic expression and identities?

An algebraic expression is an expression which consists of variables and constants. In expressions, a variable can take any value. Thus, the expression value can change if the variable values are changed. But algebraic identity is equality which is true for all the values of the variables.

How to verify algebraic identity?

The algebraic identities are verified using the substitution method. In this method, substitute the values for the variables and perform the arithmetic operation. Another method to verify the algebraic identity is the activity method. In this method, you should need a prerequisite knowledge of Geometry and some materials needed to prove the identity.

1 Comment

  1. Subhra Chakraborti

    Hi Team Byjus nice work I love reading with Byjus

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