## What are Algebraic Identities for Class 8?

• Algebraic identities are algebraic equations which are always true for every value of variables in them.

• Algebraic identities have their application in the factorization of polynomials.

• They contain variables and constants on both sides of the equation.

• In an algebraic identity, the left-side of the equation is equal to the right-side of the equation.

• For example, (a+b)2 = a2+2ab+b2 , which is true for all the values of and b.

### Using Substitution Method.

• Substitution generally means putting numbers or values in the place of variables or letters.

• In the substitution method, an arithmetic operation is performed by substituting the values for the variables.

• For example, when we have x-2=4

When we substitute x= 6,

On the Right-hand side,

4

On the left hand-side,

x-2 = 6 – 2 = 4

Here, Right hand side = Left hand side which means (x-2) is an identity.

Suppose, (a+3) (a-3) = (a2-9)

Substituting a= 1

On the Right- hand side,

(a2-9) = (1-9) = -8

On the Left- hand side,

(a+3) (a-3) = (1+3) (1-3) = (4) (-2) = -8

Here, Right hand side = Left hand side which means that (a+3) (a-3) is an identity.

### Using Activity Method.

• In this method, the algebraic identity is verified geometrically by taking different values of a x and y.

• In the activity method, the identities are verified by cutting and pasting paper.

• To verify an identity using this method, you need to have a basic knowledge of Geometry.

### Identities Class 8 –

The standard identities class 8 are derived from the Binomial Theorem. The table below consists of some Standard identities in maths class 8.

 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+c3-3abc

### Now, you Might Think What a Binomial Theorem is!

• In algebra, the Binomial Theorem is defined as a way of expanding a binomial expression raised to a large power which might be troublesome.

• A polynomial equation with just two terms generally having a plus or a minus sign in between is known as a Binomial expression.

### A small Explanation for the Above Algebraic Identities for Class 8.

For example, let us take one of the basic identities,

(a+b)2 = a2+2ab+b2, which holds for all the values of a and b.

• An identity holds true for all the values of a and b.

• We can possibly substitute one instance of one side of the equality with its other side.

• In simple words, (a+b)2 can be replaced by a2+2ab+b2 and vice versa.

• These can be used as shortcuts which make manipulating algebra easier.

### Factoring Identities

The identities listed below in the table are factoring formulas for identities of algebraic expressions class 8.

 x2-y2 = (x+y) (x-y) x3-y3 = (x-y) (x2+xy+ y2) x3 +y3 = (x+y) (x2 -xy+ y2) x4-y4 = (x2-y2) (x2 + y2)

### Three – Variable Identities –

By manipulation of the various discussed identities

entities of algebraic expressions class 8 we get these three- variable identities.

 (x+y) (x+z) (y+z) = (x+y+z) (xy+yz+xz)-xyz x2 +y2+z2          = (x+y+z)2– 2(xy+yz+xz) x3 +y3+z3             = (x+y+z)(x2 + y2 +z2 -xy-xz-yz)

## The Four Basic Identities in Maths Class 8 have Been Listed 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

### Questions to be Solved on Identities Class 8

Question 1) Find the product of (x-1) (x-1)

Solution) We need to find the product (x-1) (x-1),

(x-1) (x-1) can also be written as (x-1)2.

We know the formula for (x-1)2, expand it

(a-b)2 = a2– 2ab+b2 where a= x, b=1

(x-1)2 = x2– 2x+1

Therefore, the product of (x-1) (x-1) is x2– 2x+1

Question 2) Find the product of (x+1) (x+1) as well as the value of it using x = 2.

Solution) We need to find the product (x+1) (x+1),

(x+1) (x+1) can also be written as (x+1)2.

We know the formula for (x+1)2, expand it

(a+b)2 = a2+ 2ab+b2 where a= x, b=1

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

Putting the value of x = 2 in equation 1,

(2)2+ 2(2) +1 = 9

Therefore, the product of (x+1) (x+1) is x2+ 2x+1 and the value of the expression is 9.

Question 3) Separate the constants and the variables from the given question.

-4, 4+x, 3x+4y, -5, 4.5y, 3y2+z

Solution) Variables are the ones which include any letter such as x, y, z etc along with the numbers.

In the given question,

Constants = -4, -5

Variables = 3x+4y, 4+x, 4.5y, 3y2+z

Question 4) Find the value of

x215

x2−15

,at x = -1.

Solution) At x = -1,

x=1,x215

x=−1,x2−15

=

(1)215

(−1)2−15

= 0

Question 5) Find the value of x2+y2 – 10 at x=0 and y=0?

Solution) At x= 0 and y = 0,

x2+y2 – 10 = (0)2+(0)2 – 10

= -10

Question 6) Solve the following (x+2)2 using the concept of identities.

Solution) According to the identities and algebraic expression class 8,

We know the formula,

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

Where, a= x, b= 2

Let’s expand the given (x+2)2,

Therefore, (x+2)2 = x2+4x+4 is the solution.