Standard Algebraic Identities | Class 9 Maths

 An algebraic identity is an equality that holds for any value of its variables. They are generally used in the factorization of polynomials or simplification of algebraic calculations. Many everyday situations can be formulated in the form of mathematical equations. Identities help us by giving away to factorize them.

While operating a mobile phone. There are millions of chips under the screen of your phone, and they do their job perfectly so that you rarely feel any glitches. Thousands of engineers brainstorming on complex equations have made this possible. They deal with much bigger equations and problems. Algebraic identities then become small tools to help solve those problems. So maybe they don’t come in front of us directly, but somewhere behind the scenes, someone has definitely used them and making life easier for us.

Types of Polynomials

Based on the number of terms present in an Algebraic Expression, For example, there can be 1 term, 2 terms, 3 terms, 4 terms, and so on, they are classified into different categories. The terms are separated by either a positive or a negative sign:

Types of Polynomials	Definition	Example
Monomials	A Polynomial containing only one term	10,  2x2,  4abc
Binomials	A Polynomial containing two terms	x + y,  3p2 – 5,  x3y + 8z
Trinomials	A Polynomial containing three terms	p + q + r, x5 + 5x + 3,
Quadrinomial	A Polynomial containing four terms	p + q + r + s, m2n – 2mn + 3m + 7
Quintinomial	A Polynomial containing five terms	5x3y + 6x2 – 9xy + 8y – 7

 Note: 

• Quadrinomials and Quintinomials are not very famous terms and are often referred as Polynomials.
• The most common Standard Identities are from Binomials and Trinomials.

Standard Identities of Algebraic Expression

All standard Algebraic Identities are derived from the Binomial Theorem. There are a number of algebraic identities but a few are standard that are listed below:

Standard Identities

(a + b)2 = a2 + b2 + 2ab

(a – b)2 = a2 + b2 -2ab

a2 – b2 = (a + b)(a – b)

(ax + b)(ax – b) = ax2 – b2

(x + a) (x + b) = x2 + (a + b)x + ab

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

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Let’s look at the different applications of these identities. 

Applications of Identities

Identity 1: (a + b)² = a²  + b² + 2ab

Question: Find the value of (x + 6)(x + 6) using algebraic identities when x = 3. 

Solution:

(x+6)(x+6) can be re-written as (x + 6)². 

It can be rewritten in this form, (a + b)² = a² + b² + 2ab. 

(x + 6)² = x² + 6² + 2(6x) 

             = x² + 36 + 12x 

Given, x = 3. 

(x + 6)² = 3² + 36 + 12(3) 

             = 9 + 36 + 36 

             = 81

Identity 2: (a – b)² = a² + b²  -2ab

Question: Expand (5x – 3y)².

Solution:

This is similar to expanding (a – b)² = a² + b² – 2ab.

where a = 5x and b = 3y,

So (5x – 3y)² = (5x)² + (3y)² – 2(5x)(3y)

                    = 25x² + 9y² – 30xy 

Identity 3: a² – b² = (a + b)(a – b)

Question: Factorize (x⁶ – 1) using the identities mentioned above. 

Solution:

(x⁶ – 1) can be written as (x³)² – 1². 

This resembles the identity a² – b² = (a + b)(a – b). 

where a = x³, and b = 1. 

So, x⁶ – 1 = (x³)² – 1 = (x³ + 1) (x³ – 1). 

Solved Examples on Standard Algebraic Identities

Question 1: If a + b = 12 and ab = 35, what is a⁴ + b⁴? 

Solution:

a⁴ + b⁴ can be written as (a²)² + (b²)², 

And we know, (x + y)² = x² + y² + 2xy

⇒ x² + y² = (x + y)² -2xy 

So, in this case, x = a², y = b² ;

a⁴ + b⁴ = (a² + b²)² – 2(a²)(b²) 

⇒ ((a+b)² – 2ab)² – 2(a²)(b²) 

⇒ ((12)² – 2(35))² – 2(35)²

⇒ 5475 – 2450

⇒ 3026

Question 2: The identity 4(z + 7)(2z – 1) = Az² + Bz + C holds for all real values of z. What is A + B + C?

Solution:

Multiplying out the left side of the identity, we have

4(x + 7)(2x − 1) = 8x² + 52x − 28.

This expression must be equal to the right-hand side of the identity, implying

8x² + 52x – 28 = Ax² + Bx + C,

So now comparing both sides of the equation. 

A = 8, B = 52 ad C – 28. 

A + B + C  = 8 + 52 – 28 = 32

Question 3: If a + b + c = 6, a² + b² + c² = 14 and ab + bc + ca = 11, what is a³ + b³ + c³ – 3abc?

Solution:

We know this identity,

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

Substituting the given values, 

a³ + b³ + c³ -3abc = (6)(14 -11)

⇒ (6)(3) = 18

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