# NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities- As you have learned in previous classes that expressions are formed from variables and constants. This chapter will introduce you to the application of algebraic terms and variables to solve various problems. NCERT solutions for Class 8 Maths chapter 9 Algebraic Expressions and Identities will give you a detailed explanation for every problem of this chapter. Algebra is the most important branch of mathematics which teaches how to form equations and solving them using different kinds of techniques. Important topics like the product of the equation, finding the coefficient of the variable in the equation, subtraction of the equation and creating quadratic equation by the product of its two roots, and division of the equation are covered in this chapter.

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You will find the questions related to these topics in NCERT solutions for Class 8 Maths chapter 9 Algebraic Expressions and Identities which will make your task easy while solving the problems. As it is a new concept, you may find difficulties while dealing with algebraic parts of mathematics but if you practice questions and go through NCERT solutions for Class 8 Maths chapter 9 Algebraic Expressions and Identities, you will find it very easy and one of the strongest parts in Mathematics. You will get NCERT Solutions from Classes 6 to 12 for Science and Maths by clicking on the above link. Here you will get the detailed NCERT Solutions for Class 8 by clicking on the link.

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## what are expressions?

Question: 1 Give five examples of expressions containing one variable and five examples of expressions containing two variables.

Five examples of expressions containing one variable are:

Five examples of expressions containing two variables are:

Question: 2(i) Show on the number line

x on the number line:

Question: 2(ii) Show on the number line :

x-4 on the number line:

Question: 2(iii) Show on the number line :

2x+1

2x+1 on the number line:

Question: 2(iv) Show on the number line:

3x - 2 on the number line

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Topic 9.2 Terms, Factors and Coefficients

Question:1 Identify the coefficient of each term in the expression.

coefficient of each term are given below

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Topic 9.3 Monomials, Binomials and Polynomials

Question: 1(i) Classify the following polynomials as monomials, binomials, trinomials.

Binomial since there are two terms with non zero coefficients.

Question: 1(ii) Classify the following polynomials as monomials, binomials, trinomials.

Trinomial since there are three terms with non zero coefficients.

Question:1(iii) Classify the following polynomials as monomials, binomials, trinomials.

Trinomial since there are three terms with non zero coefficients.

Question: 1(iv) Classify the following polynomials as monomials, binomials, trinomials.

Binomial since there are two terms with non zero coefficients.

Question: 1(v) Classify the following polynomials as monomials, binomials, trinomials.

Monomial since there is only one term.

Question: 2(a) Construct 3 binomials with only as a variable;

Three binomials with the only x as a variable are:

Question: 2(b) Construct 3 binomials with and as variables;

Three binomials with x and y as variables are:

Question: 2(c) Construct 3 monomials with and as variables;

Three monomials with x and y as variables are

Question: 2(d) Construct 2 polynomials with 4 or more terms .

Two polynomials with 4 or more terms are:

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Topic 9.4 Like and Unlike Terms

Question:(i) Write two terms which are like

Question:(ii) Write two terms which are like

we can write more like terms

Question:(iii) Write two terms which are like

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities-Exercise: 9.1

Question:1(i) Identify the terms, their coefficients for each of the following expressions.

following are the terms and coefficient

The terms are and the coefficients are 5 and -3.

Question: 1(ii) Identify the terms, their coefficients for each of the following expressions.

the following is the solution

Question:1(iii) Identify the terms, their coefficients for each of the following expressions.

Question: 1(iv) Identify the terms, their coefficients for each of the following expressions.

The terms are 3, -pq, qr,and -rp and the coefficients are 3, -1, 1 and -1 respectively.

Question:1(v) Identify the terms, their coefficients for each of the following expressions.

Above are the terms and coefficients

Question: 1(vi) Identify the terms, their coefficients for each of the following expressions.

The terms are 0.3a, -0.6ab and 0.5b and the coefficients are 0.3, -0.6 and 0.5.

Question: 2(a) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Binomial.

Question: 2(b) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Monomial.

Question: 2(c) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

This polynomial does not fit in any of these three categories.

Question: 2(d) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Trinomial.

Question: 2(e) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Binomial.

Question: 2(f) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Trinomial.

Question: 2(g) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Trinomial.

Question: 2(h) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Binomial.

Question: 2(i) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

This polynomial does not fit in any of these three categories.

Question:2(j) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Monomial.

Question: 2(k) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Binomial.

Question: 2(i) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Binomial.

ab-bc+bc-ca+ca-ab=0.

Question: 4(a) Subtract from

12a-9ab+5b-3-(4a-7ab+3b+12)
=(12-4)a +(-9+7)ab+(5-3)b +(-3-12)
=8a-2ab+2b-15

Question: 4(b) Subtract from

Question: 4(c) Subtract from

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Topic 9.7.2 Multiplying Three or More Monomials

Question:1 Find . First find and multiply it by ; or first find and multiply it by .

We observe that the result is same in both cases and the result does not depend on the order in which multiplication has been carried out.

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities-Exercise: 9.2

Question: 1(i) Find the product of the following pairs of monomials.

Question: 1(ii) Find the product of the following pairs of monomials.

Question: 1(iii) Find the product of the following pairs of monomials

Question: 1(iv) Find the product of the following pairs of monomials.

Question:1(v) Find the product of the following pairs of monomials.

Question:2(A) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

The question can be solved as follows

Question:2(B) Find the areas of rectangles with the following pairs of monomials as their lengths and breadth respectively.

the area is calculated as follows

Question:2(C) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

the following is the solution

Question:2(D) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

area of rectangles is

Question:2(E) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

The area is calculated as follows

Question:3 Complete the table of products.

 First monomial Second monomial ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...

 First monomial Second monomial

Question:4(i) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

Question:4(ii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

the volume of rectangular boxes with the following length, breadth and height is

Question:4(iii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

the volume of rectangular boxes with the following length, breadth and height is

Question:4(iv) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

the volume of rectangular boxes with the following length, breadth and height is

Question:5(i) Obtain the product of

the product

Question:5(ii) Obtain the product of

the product

Question:5(iii) Obtain the product of

the product

Question:5(iv) Obtain the product of

the product

Question:5(v) Obtain the product of

the product

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Topic 9.8.1 Multiplying a Monomial by a Binomial

Question:(i) Find the product

Using distributive law,

Question:(ii) Find the product

Using distributive law,

We have :

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Topic 9.8.2 Multiplying A Monomial By A Trinomial

Question:1 Find the product:

By using distributive law,

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities-Exercise: 9.3

Question:1(i) Carry out the multiplication of the expressions in each of the following pairs.

Multiplication of the given expression gives :

By distributive law,

Question:1(ii) Carry out the multiplication of the expressions in each of the following pairs.

We have ab, (a-b).

Using distributive law we get,

Question:1(iii) Carry out the multiplication of the expressions in each of the following pairs.

Using distributive law we can obtain multiplication of given expression:

Question:1(iv) Carry out the multiplication of the expressions in each of the following pairs.

We will obtain multiplication of given expression by using distributive law :

Question:1(v) Carry out the multiplication of the expressions in each of the following pairs.

Using distributive law :

Question:2 Complete the table

 First expression Second expression Product (i) ... (ii) ... (iii) ... (iv) ... (v) ...

We will use distributive law to find product in each case.

 First expression Second expression Product (i) (ii) (iii) (iv) (v)

Question:3(i) Find the product.

Opening brackets :

or

Question:3(ii) Find the product.

We have,

Question:3(iii) Find the product.

We have

Question:3(iv) Find the product.

We have

or

Question:4(a) Simplify and find its values for

(i)

(a) We have

Put x = 3,

We get :

Question:4(a) Simplify and find its values for

(ii)

We have

Put

. So We get,

Question:4(b) Simplify and find its value for

(i)

We have :

Put a = 0 :

Question:4(b) Simplify and find its value for

(ii)

We have

Put a = 1 ,

we get :

Question:4(b) Simplify and find its value for

(iii)

We have .

or

Put a = (-1)

(a)First we will solve each brackets individually.

; ;

Firstly, open the brackets:

and

or

Question:5(c) Subtract: from

At first we will solve each bracket individually,

and

Subtracting:

or

or

Question:5(d) Subtract: from

Solving brackets :

and

Subtracting :

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities-Exercise: 9.4

Question:1(i) Multiply the binomials.

and

We have (2x + 5) and (4x - 3)
(2x + 5) X (4x - 3) = (2x)(4x) + (2x)(-3) + (5)(4x) + (5)(-3)
= 8 - 6x + 20x - 15
= 8 + 14x -15

Question:1(ii) Multiply the binomials.

and

We need to multiply (y - 8) and (3y - 4)
(y - 8) X (3y - 4) = (y)(3y) + (y)(-4) + (-8)(3y) + (-8)(-4)
= 3 - 4y - 24y + 32
= 3 - 28y + 32

Question:1(iii) Multiply the binomials

and

We need to multiply (2.5l - 0.5m) and (2.5l + 0..5m)
(2.5l - 0.5m) X (2.5l + 0..5m) = using
= 6.25 - 0.25

Question:1(iv) Multiply the binomials.

and

(a + 3b) X (x + 5) = (a)(x) + (a)(5) + (3b)(x) + (3b)(5)
= ax + 5a + 3bx + 15b

Question:1(v) Multiply the binomials.

and

(2pq + 3q 2 ) X (3pq - 2q 2 ) = (2pq)(3pq) + (2pq)(-2q 2 ) + ( 3q 2 )(3pq) + (3q 2 )(-2q 2 )
= 6p 2 q 2 - 4pq 3 + 9pq 3 - 6q 4
= 6p 2 q 2 +5pq 3 - 6q 4

Question:1(vi) Multiply the binomials.

and

Multiplication can be done as follows

X =

=

=

Question:2(i) Find the product.

(5 - 2x) X (3 + x) = (5)(3) + (5)(x) +(-2x)(3) + (-2x)(x)
= 15 + 5x - 6x - 2
= 15 - x - 2

Question:2(ii) Find the product.

(x + 7y) X (7x - y) = (x)(7x) + (x)(-y) + (7y)(7x) + (7y)(-y)
= 7 - xy + 49xy - 7
= 7 + 48xy - 7

Question:2(iii) Find the product.

( + b) X (a + ) = ( )(a) + ( )( ) + (b)(a) + (b)( )
=

Question:2(iv) Find the product.

following is the solution

( ) X (2p + q) =

Question:3(i) Simplify.

this can be simplified as follows

( -5) X (x + 5) + 25 = ( )(x) + ( )(5) + (-5)(x) + (-5)(5) + 25
=
=

Question:3(ii) Simplify .

This can be simplified as

( + 5) X ( + 3) + 5 = ( )( ) + ( )(3) + (5)( ) + (5)(3) + 5
=
=

Question:3(iii) Simplify.

simplifications can be

(t + )( - s) = (t)( ) + (t)(-s) + ( )( ) + ( )(-s)
=

Question:3(iv) Simplify.

(a + b) X ( c -d) + (a - b) X (c + d) + 2(ac + bd )
= (a)(c) + (a)(-d) + (b)(c) + (b)(-d) + (a)(c) + (a)(d) + (-b)(c) + (-b)(d) + 2(ac + bd )
= ac - ad + bc - bd + ac +ad -bc - bd + 2(ac + bd )
= 2(ac - bd ) + 2(ac +bd )
= 2ac - 2bd + 2ac + 2bd
= 4ac

Question:3(v) Simplify.

(x + y) X ( 2x + y) + (x + 2y) X (x - y)
=(x)(2x) + (x)(y) + (y)(2x) + (y)(y) + (x)(x) + (x)(-y) + (2y)(x) + (2y)(-y)
= 2 + xy + 2xy + + - xy + 2xy - 2
=3 + 4xy -

Question:3(vi) Simplify.

simplification is done as follows

(x + y) X ( ) = x X ( ) + y ( )
=
=

Question:3(vii) Simplify.

(1.5x - 4y) X (1.5x + 4y + 3) - 4.5x + 12y = (1.5x) X (1.5x + 4y + 3) -4y X (1.5x + 4y + 3) - 4.5x + 12y
= 2.25 + 6xy + 4.5x - 6xy - 16 - 12y -4.5x + 12 y
= 2.25 - 16

Question:3(viii) Simplify.

(a + b + c) X (a + b - c) = a X (a + b - c) + b X (a + b - c) + c X (a + b - c)
= + ab - ac + ab + -bc + ac + bc -
= + - + 2ab

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities Topic 9.11 Standard Identities

Question:1(i) Put -b in place of b in identity 1. Do you get identity 2?

Identity 1
If we replace b with -b in identity 1
We get,

which is equal to
which is identity 2
So, we get identity 2 by replacing b with -b in identity 1

## NCERT Free Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities Topic 9.11 Standard Identities

Question:1 Verify Identity (IV), for .

Identity IV
(a + x)(b + x) =
So, it is given that a = 2, b = 3 and x = 5
Lets put these value in identity IV
(2 + 5)(3 + 5) = + (2 + 3)5 +2 X 3
7 X 8 = 25 + 5 X 5 + 6
56 = 25 + 25 + 6
= 56
L.H.S. = R.H.S.
So, by this we can say that identity IV satisfy with given value of a,b and x

Question:2 Consider, the special case of Identity (IV) with a = b, what do you get? Is it related to Identity

Identity IV is
If a =b than

(a + x)(a + x) =

Which is identity I

Question:3 Consider, the special case of Identity (IV) with and What do you get? Is it related to Identity ?

Identity IV is
If a = b = -c than,
(x - c)(x - c) =

Which is identity II

Question:4 Consider the special case of Identity (IV) with . What do you get? Is it related to Identity?

Identity IV is
If b = -a than,

(x + a)(x - a) =
=
Which is identity III

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities-Exercise: 9.5

Question:1(i) Use a suitable identity to get each of the following products.

(x + 3) X (x +3) =
So, we use identity I for this which is

In this a=x and b = x

=

Question:1(ii) Use a suitable identity to get each of the following products in bracket.

(2y + 5) X ( 2y + 5) =
We use identity I for this which is

IN this a = 2y and b = 5

=

Question:1(iii) Use a suitable identity to get each of the following products in bracket.

(2a -7) X (2a - 7) =
We use identity II for this which is

in this a = 2a and b = 7

=

Question:1(iv) Use a suitable identity to get each of the following products in bracket.

We use identity II for this which is

in this a = 3a and b = -1/2

=

Question:1(v) Use a suitable identity to get each of the following products in bracket.

We use identity III for this which is
(a - b)(a + b) =
In this a = 1.1m and b = 4
=
= 1.21 - 16

Question:1(vi) Use a suitable identity to get each of the following products in bracket.

take the (-)ve sign common so our question becomes
-
We use identity III for this which is
(a - b)(a + b) =
In this a = and b =

=

Question:1(vii) Use a suitable identity to get each of the following.

(6x -7) X (6x - 7) =
We use identity III for this which is
(a - b)(a + b) =
In this a = 6x and b = 7
(6x -7) X (6x - 7) =

Question:1(viii) Use a suitable identity to get each of the following product.

take (-)ve sign common from both the brackets So, our question become
(a -c) X (a -c) =
We use identity II for this which is

In this a = a and b = c

Question:1(ix) Use a suitable identity to get each of the following product.

We use identity I for this which is

In this a = and b =

=

Question:1(x) Use a suitable identity to get each of the following products.

We use identity II for this which is

In this a = 7a and b = 9b

=

Question:2(i) Use the identity to find the following products.

We use identity
in this a = 3 and b = 7
=
=

Question:2(ii) Use the identity to find the following products.

We use identity
In this a= 5 , b = 1 and x = 4x
=
=

Question:2(iii) Use the identity to find the following products.

We use identity
in this x = 4x , a = -5 and b = -1
=
=

Question:1(iv) Use the identity to find the following products.

We use identity
In this a = 5 , b = -1 and x = 4x
=
=

Question:2(v) Use the identity to find the following products.

We use identity
In this a = 5y , b = 3y and x = 2x
=
=

Question:2(vi) Use the identity to find the following products.

We use identity
In this a = 9 , b = 5 and x =
=
=

Question:2(vii) Use the identity to find the following products.

We use identity
In this a = -4 , b = -2 and x = xyz
=
=

Question:3(i) Find the following squares by using the identities.

We use identity

In this a =b and b = 7

=

Question:3(ii) Find the following squares by using the identities.

We use

In this a = xy and b = 3z

=

Question:3(iii) Find the following squares by using the identities.

We use

In this a = and b =

=

Question:3(iv) Find the following squares by using the identities.

we use the identity

In this a = and b =

=

Question:3(v) Find the following squares by using the identities.

we use

In this a = 0.4p and b =0.5q

=

Question:3(vi) Find the following squares by using the identities.

we use the identity

In this a = 2xy and b =5y

=

Question:4(i) Simplify:

we use

In this a = and b =

=

Question:4(ii) Simplify.

we use

In this a = (2x + 5) and b = (2x - 5)

=
= (4x)(10)
=40x

or

remember that

here a= 2x, b= 5

Question:4(iii) Simplify.

we use
and
In this a = 7m and b = 8n

=
and

=

So, = +
=

or

remember that

Question: 4(iv) Simplify.

we use

1 ) In this a = 4m and b = 5n

=
2 ) in this a = 5m and b = 4n

=

So, = +
=

Question: 4(v) Simplify.

we use

1 ) In this a = (2.5p- 1.5q) and b = (1.5p - 2.5q)

=
= 4(p + q ) (p - q)
= 4

Question:4(vi) Simplify.

We use identity

In this a = ab and b = bc

=
Now, -
=

Question:4(vii) Simplify.

We use identity

In this a = and b =

=
Now, +
=

Question:5(i) Show that

L.H.S. =

= R.H.S.

Hence it is prooved

Question:5(ii) Show that

L.H.S. = (Using )

= R.H.S.

Question:5(iii) Show that.

First we will solve the LHS :

or

= RHS

Question:5(iv) Show that.

Opening both brackets we get,

= R.H.S.

Question:5(v) Show that

Opening all brackets from the LHS, we get :

= RHS

Question:6(i) Using identities, evaluate.

We will use the identity:

So,

Question:6(ii) Using identities, evaluate.

Here we will use the identity :

So :

or

Question:6(iii) Using identities, evaluate.

Here we will use the identity :

So :

or

Question:6(iv) Using identities, evaluate.

Here we will the identity :

or

or

Question:6(v) Using identities, evaluate.

Here we will use :

Thus

or

Question:6(vi) Using identities, evaluate.

This can be written as :

using

or

Question:6(vii) Using identities, evaluate.

This can be written in form of :

or

or

Question:6(viii) Using identities, evaluate.

Here we will use the identity :

Thus :

or

or

Question:6(ix) Using identities, evaluate.

This can be written as :

or

or

or

Question:7(i) Using , find

We know,

Using this formula,

= (51 + 49)(51 - 49)

= (100)(2)

= 200

Question:7(ii) Using , find

We know,

Using this formula,

= (1.02 + 0.98)(1.02 - 0.98)
= (2.00)(0.04)

= 0.08

Question:7(iii) Using , find.

We know,

Using this formula,

= (153 - 147)(153 +147)

=(6) (300)

= 1800

Question:7(iv) Using , find

We know,

Using this formula,

= (1.02 + 0.98)(1.02 - 0.98)

= (2.00)(0.04)

= 0.08

Question:8(i) Using

We know,

Using this formula,

= (100 + 3)(100 + 4)

Here x =100, a = 3, b = 4

= 11212

Question:8(ii) Using , find

We know,

Using this formula,

= (5 + 0.1)(5 + 0.2)

Here x =5, a = 0.1, b = 0.2

= 26.52

Question:8(iii) Using , find

We know,

Using this formula,

= (100 + 3)(100 - 2) = (100 + 3){100 + (-2)}

Here x =100, a = 3, b = -2

= 10094

Question: 8(iv) Using , find

We know,

Using this formula,

= (10 - 0.3)(10 - 0.2) = {10 + (-0.3)}{10 + (-0.2)}

Here x =10, a = -0.3, b = -0.2

= 95.

## Algebraic Expressions and Identities Class 8 Math Chapter 9-Topics

• What are Expressions?
• Terms, Factors and Coefficients
• Monomials, Binomials,f and Polynomials
• Like and Unlike Terms
• Addition and Subtraction of Algebraic Expressions
• Multiplication of Algebraic Expressions: Introduction
• Multiplying a Monomial by a Monomial
• Multiplying two monomials
• Multiplying three or more monomials
• Multiplying a Monomial by a Polynomial
• Multiplying a monomial by a binomial
• Multiplying a monomial by a trinomial
• Multiplying a Polynomial by a Polynomial
• Multiplying a binomial by a binomial
• Multiplying a binomial by a trinomial
• What is an Identity?
• Standard Identities
• Applying Identities

### NCERT Solutions for Class 8 Maths: Chapter-Wise

 Chapter -1 Rational Numbers Chapter -2 Linear Equations in One Variable Chapter-3 Understanding Quadrilaterals Chapter-4 Practical Geometry Chapter-5 Data Handling Chapter-6 Squares and Square Roots Chapter-7 Cubes and Cube Roots Chapter-8 Comparing Quantities Chapter-9 Algebraic Expressions and Identities Chapter-10 Visualizing Solid Shapes Chapter-11 Mensuration Chapter-12 Exponents and Powers Chapter-13 Direct and Inverse Proportions Chapter-14 Factorization Chapter-15 Introduction to Graphs Chapter-16 Playing with Numbers

### NCERT Solutions for Class 8: Subject-Wise

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science

Some Important Identities From NCERT Book for Class 8 Chapter 9 Algebraic Expressions And Identities

you can write

Can be simplified as follows