# NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5

## Introduction:

In this exercise/article we will learn about multiplication of algebraic expressions. [ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} and ( a - b )^{2} = a^{2} - 2ab + b^{2} and ( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab and ( a - b ) ( a + b ) = a^{2} - b^{2} you learn identity before then do this exercise. In all question put on identity.

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities :

- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.1
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.2
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.3
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.4
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.5

**Class 8 Maths Exercise 9.5 (Page-151)**

**Q1. **Use a suitable identity to get each of the following products.

**(i) **( x + 3 ) ( x + 3 )

**Solution :**

We have ( x + 3 ) ( x + 3 ) = ( x + 3 )^{2}

= ( x + 3 )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here a = x and b = 3

So that,

( x )^{2} + 2 × x × 3 + ( 3 )^{2}

= x^{2} + 6x + 9

So, answer is x^{2} + 6x + 9

**(ii)** ( 2y + 5 ) ( 2y + 5 )

**Solution :**

We have ( 2y + 5 ) ( 2y + 5 ) = ( 2y + 5 )^{2}

= ( 2y + 5 )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here, a = 2y and b = 5

So that,

( 2y )^{2} + 2 × 2y × 5 + ( 5 )^{2}

= 4y^{2} + 20y + 25

So, answer is 4y^{2} + 20y + 25

**(iii)** ( 2a - 7 ) ( 2a - 7 )

**Solution :**

We have ( 2a - 7 ) ( 2a - 7 ) = ( 2a - 7 )^{2}

= ( 2a - 7 )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here, a = 2a and b = 7

So that,

( 2a )^{2} - 2 × 2a × 7 + ( 7 )^{2}

= 4a^{2} - 28a + 49

So, answer is 4a^{2} - 28a + 49

**(iv)** \(\displaystyle (\,3a\,-\,\frac{1}{2}\,)\,(\,3a\,-\,\frac{1}{2}\,)\)

**Solution : **

We have \(\displaystyle (\,3a\,-\,\frac{1}{2}\,)\,(\,3a\,-\,\frac{1}{2}\,)\)\(\displaystyle ={{(\,3a\,-\,\frac{1}{2}\,)}^{2}}\)

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here, a = 3a and b = \(\displaystyle \frac{1}{2}\)

So that,

( 3a )^{2} - 2 × 3a × \(\displaystyle \frac{1}{2}\) + ( \(\displaystyle \frac{1}{2}\) )^{2}

= 9a^{2} - 3a + \(\displaystyle \frac{1}{4}\)

So, answer is 9a^{2} - 3a + \(\displaystyle \frac{1}{4}\)

**(v) **( 1.1m - 0.4 ) ( 1.1m + 0.4 )

We have ( 1.1m - 0.4 ) ( 1.1m + 0.4 )

[ using identity = ( a - b ) ( a + b ) = a^{2} - b^{2} ]

Here, a = 1.1m and b = 0.4

( 1.1m )^{2} - ( 0.4 )^{2}

= 1.21m^{2} - 0.16

So, answer is 1.21m^{2} - 0.16

**(vi) **( a^{2} + b^{2} ) ( - a^{2} + b^{2} )

**Solution :**

We have ( a^{2} + b^{2} ) ( - a^{2} + b^{2} )

[ we have multiply to the first bracket by the second bracket ]

a^{2 }( - a^{2} + b^{2} ) + b^{2} ( - a^{2} + b^{2} )

= - a^{4} + a^{2}b^{2} - a^{2}b^{2} + b^{4} [ because ( a^{2}b^{2} - a^{2}b^{2} ) = 0 ]

= b^{4} - a^{4}

So, answer is b^{4} - a^{4}

**(vii) **( 6x - 7 ) ( 6x + 7 )

**Solution :**

We have ( 6x - 7 ) ( 6x + 7 )

[ using identity = ( a - b ) ( a + b ) = a^{2} - b^{2} ]

Here, a = 6x and b = 7

So that,

( 6x )^{2} - ( 7 )^{2}

= 36x^{2} - 49

So, answer is 36x^{2} - 49

**(viii) **( - a + c )** **( - a + c )

**Solution :**

We have ** **( - a + c )** **( - a + c ) = ( - a + c )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here, a = - a and b = c

So that,

( - a )^{2} + 2 × - a × c + ( c )^{2}

= a^{2} - 2ac + c^{2}

So, answer is a^{2} - 2ac + c^{2}

**(ix) **\(\displaystyle \left( {\frac{x}{2}\,+\,\frac{{3y}}{4}} \right)\) \(\displaystyle \left( {\frac{x}{2}\,+\,\frac{{3y}}{4}} \right)\)

**Solution :**

We have \(\displaystyle \left( {\frac{x}{2}\,+\,\frac{{3y}}{4}} \right)\) \(\displaystyle \left( {\frac{x}{2}\,+\,\frac{{3y}}{4}} \right)\) = \(\displaystyle {{\left( {\frac{x}{2}+\frac{{3y}}{4}} \right)}^{2}}\)

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here, a = \(\displaystyle {\frac{x}{2}}\) and b = \(\displaystyle {\frac{{3y}}{4}}\)

So that,

= \(\displaystyle {{\left( {\frac{x}{2}} \right)}^{2}}\,+\,2\,\times \,\frac{x}{2}\,\times \,\frac{{3y}}{4}\,+\,{{\left( {\frac{{3y}}{4}} \right)}^{2}}\)

=

= \(\displaystyle \frac{{{{x}^{2}}}}{4}\,+\,\frac{{3xy}}{4}\,+\,\frac{{9{{y}^{2}}}}{{16}}\)

So, answer is \(\displaystyle \frac{{{{x}^{2}}}}{4}\,+\,\frac{{3xy}}{4}\,+\,\frac{{9{{y}^{2}}}}{{16}}\)

**(x) **( 7a - 9b ) ( 7a - 9b )

**Solution :**

We have ( 7a - 9b ) ( 7a - 9b ) = ( 7a - 9b )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here, a = 7a and b = 9b

So that,

( 7a )^{2} - 2 × 7a × 9b + ( 9b )^{2}

= 49a^{2} - 126ab + 81b^{2}

So, answer is 49a^{2} - 126ab + 81b^{2}

**Q2. **Use the identity ( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab to find the following products.

**(i)** ( x + 3 ) ( x + 7 )

**Solution :**

We have ( x + 3 ) ( x + 7 )

[ using identity = ( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

Here, x = x , a = 3 and b = 7

So that,

( x )^{2} + ( 3 + 7 ) x + 3 × 7

= x^{2 }+ ( 10 ) x + 21

= x^{2 }+ 10 × x + 21

= x^{2 }+10x +21

So, answer is x^{2 }+10x +21

**(ii) **( 4x + 5 ) ( 4x + 1 )

**Solution :**

We have ( 4x + 5 ) ( 4x + 1 )

[ using identity = [( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

Here, x = 4x , a = 5 and b = 1

So that,

( 4x )^{2} + ( 5 + 1 ) 4x + 5 × 1

= 16x^{2} + ( 6 ) 4x + 5

= 16x^{2} + 6 × 4x + 5

= 16x^{2} + 24x + 5

So, answer is 16x^{2} + 24x + 5

**(iii) **( 4x - 5 ) ( 4x - 1 )

We have ( 4x - 5 ) ( 4x - 1 )

[ using identity = [( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

Here, x = 4x , a = - 5 and b = -1

So that,

[( 4x )^{2} + {( - 5 ) + ( -1 )} 4x + ( - 5 ) × ( - 1 )]

= 16x^{2} + (- 6) 4x + ( 5 )

= 16x^{2} - 6 × 4x + 5

= 16x^{2} - 24x + 5

So, answer is 16x^{2} - 24x + 5

**(iv) **( 4x + 5 ) ( 4x - 1 )

**Solution :**

We have ( 4x + 5 ) ( 4x - 1 )

[ using identity = [( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

Here, x = 4x , a = 5 and b = - 1

So that,

[( 4x )^{2} + { 5 + ( -1 ) } 4x + 5 × ( - 1 )]

= 16x^{2} ( 4 ) 4x - 5

= 16x^{2} + 4 × 4x - 5

= 16x^{2} + 16x - 5

So, answer is 16x^{2} + 16x - 5

**(v) **( 2x + 5y ) ( 2x + 3y )

**Solution :**

We have ( 2x + 5y ) ( 2x + 3y )

[ using identity = [( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

Here, x = 2x , a = 5y and b = 3y

So that,

( 2x )^{2} + ( 5y + 3y ) 2x + 5y × 3y

= 4x^{2} + ( 8y ) 2x + 15y^{2}

= 4x^{2} + 8y × 2x + 15y^{2}

= 4x^{2} + 16xy + 15y^{2}

So, answer is 4x^{2} + 16xy + 15y^{2}

**(vi) **( 2a^{2} + 9 ) ( 2a^{2} + 5 )

We have ( 2a^{2} + 9 ) ( 2a^{2} + 5 )

[ using identity = [( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

Here, x = 2a^{2} , a = 9 and b = 5

So that,

( 2a )^{2} + ( 9 + 5 ) 2a^{2} + 9 × 5

= 4a^{4} + ( 14 ) 2a^{2} + 45

= 4a^{4} + 14 × 2a^{2} + 45

= 4a^{4} + 28a^{2} + 45

So, answer is 4a^{4} + 28a^{2} + 45

**(vii) **( xyz - 4 ) ( xyz - 2 )

**Solution :**

We have ( xyz - 4 ) ( xyz - 2 )

[ using identity = [( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

Here, x = xyz , a = - 4 and b = - 2

So that,

[( xyz )^{2} + { ( - 4 ) + ( - 2 ) } xyz + ( - 4 × - 2 )]

= x^{2}y^{2}z^{2} + ( - 6 ) xyz + 8

= x^{2}y^{2}z^{2} - 6 × xyz + 8

= x^{2}y^{2}z^{2} - 6xyz + 8

So, answer is x^{2}y^{2}z^{2} - 6xyz + 8

**Q3. **Find the following squares by using the identities.

(i) ( b - 7 )^{2}

**Solution :**

We have ( b - 7 )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here , a = b and b = 7

So that,

b^{2} - 2 × b × 7 + 7^{2}

= b^{2} - 14b + 49

So, answer is b^{2} - 14b + 49

**(ii) **( xy + 3z )^{2}

**Solution :**

We have ( xy + 3z )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here , a = xy and b = 3z

So that,

( xy )^{2} + 2 × xy × 3z + ( 3z )^{2}

= xy^{2} + 6xyz + 9z^{2}

So, answer is xy^{2} + 6xyz + 9z^{2}

**(iii) **( 6x^{2} - 5y )^{2}

**Solution :**

We have ( 6x^{2} - 5y )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here , a = 6x^{2} and b = 5y

So that,

( 6x^{2} ) - 2 × 6x^{2} × 5y + ( 5y )^{2}

= 36x^{2} - 60x^{2}y + 5y^{2}

So, answer is 36x^{2} - 60x^{2}y + 5y^{2}

**(iv)** \(\displaystyle {{\left( {\frac{2}{3}m\,+\,\frac{3}{2}n} \right)}^{2}}\)

**Solution :**

We have \(\displaystyle {{\left( {\frac{2}{3}m\,+\,\frac{3}{2}n} \right)}^{2}}\)

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here , a = \(\displaystyle {\frac{2}{3}m}\) and b = \(\displaystyle {\frac{3}{2}n}\)

So that,

= \(\displaystyle {{\left( {\frac{2}{3}m} \right)}^{2}}\,+\,2\,\times \,\frac{2}{3}m\,\times \,\frac{3}{2}n\,+\,{{\left( {\frac{3}{2}n} \right)}^{2}}\)

=

= \(\displaystyle \frac{4}{9}{{m}^{2}}\,+\,2mn\,+\,\frac{9}{4}{{n}^{2}}\)

So, answer is \(\displaystyle \frac{4}{9}{{m}^{2}}\,+\,2mn\,+\,\frac{9}{4}{{n}^{2}}\)

**(v) **( 0.4p - 0.5q )^{2}

**Solution :**

We have ( 0.4p - 0.5q )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here , a = 0.4p and b = 0.5q

So that,

( 0.4p )^{2} + 2 × 0.4p × 0.5q + ( 0.5q )^{2}

= 0.16p^{2} + 0.4pq + 0.25q^{2}

So, answer is 0.16p^{2} + 0.4pq + 0.25q^{2}

**(vi) **( 2xy + 5y )^{2}

**Solution :**

We have ( 2xy + 5y )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here , a = 2xy and b = 5y

So that,

( 2xy )^{2} + 2 × 2xy × 5y + ( 5y )^{2}

= 4x^{2}y^{2} + 20xy^{2} + 25y^{2}

So, answer is 4x^{2}y^{2} + 20xy^{2} + 25y^{2}

## Class 8 Maths Chapter 9 Exercise 9.5

**Q4. **Simplify.

**(i) **( a^{2} - b^{2} )^{2}

Solution :

We have ( a^{2} - b^{2} )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here , a = a^{2} and b = b^{2}

So that,

( a^{2} )^{2} - 2 × a^{2} × b^{2} + ( b^{2} )^{2}

= a^{4} - 2a^{2}b^{2} + b^{4}

So, answer is a^{4} - 2a^{2}b^{2} + b^{4}

**(ii) **( 2x + 5 )^{2} - ( 2x - 5 )^{2}

**Solution :**

We have ( 2x + 5 )^{2} - ( 2x - 5 )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} and ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here , a = 2x and b = 5 and a = 2x and b = 5

So that,

[ { ( 2x )^{2} + 2 × 2x × 5 + ( 5 )^{2} } - { ( 2x )^{2} - 2 × 2x × 5 + ( 5 )^{2} } ]

= ( 2x )^{2} + 2 × 2x × 5 + ( 5 )^{2} - ( 2x )^{2} + 2 × 2x × 5 - ( 5 )^{2}

= 4x^{2} + 20x + 25 - 4x^{2} + 20x - 25 [ because = 4x^{2} - 4x^{2} = 0 and 25 - 25 = 0 ]

= 40x

So, answer is 40x

**(iii) **( 7m - 8n )^{2} + ( 7m + 8n )^{2}

**Solution :**

We have ( 7m - 8n )^{2} + ( 7m + 8n )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} and ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here , a = 7m and b = 8n and a = 7m and b = 8n

So that,

[ { ( 7m )^{2} - 2 × 7m × 8n + ( 8n )^{2} } +{ ( 7m )^{2} + 2 × 7m × 8n + ( 8n )^{2} } ]

= ( 7m )^{2} - 2 × 7m × 8n + ( 8n )^{2} + ( 7m )^{2} + 2 × 7m × 8n + ( 8n )^{2}

= 49m^{2} - 112mn + 64n^{2} + 49m^{2} + 112mn + 64n^{2} [ because = - 112mn + 112mn = 0 ]

= ( 49m^{2} + 49m^{2} ) ( 64n^{2} + 64n^{2} )

= 98m^{2} + 128n^{2}

So, answer is 98m^{2} + 128n^{2}

**(vi) **( 4m + 5n )^{2} + ( 5m + 4n )^{2}

**Solution : **

We have ** **( 4m + 5n )^{2} + ( 5m + 4n )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} and ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here , a = 4m and b = 5n and a = 5m and b = 4n

So that,

( 4m )^{2} + 2 × 4m × 5n + ( 5n )^{2} + ( 5m )^{2} + 2 × 5m × 4n + ( 4n )^{2}

= 16m^{2} + 40mn + 25n^{2} + 25m^{2} + 40mn + 16n^{2}

= ( 16m^{2} + + 25m^{2} ) ( 40mn + 40mn ) ( 25n^{2} + 16n^{2} )

= 41m^{2} + 80mn + 41n^{2}

So, answer is 41m^{2} + 80mn + 41n^{2}

**(v) **( 2.5p - 1.5q )^{2} - ( 1.5p - 2.5q )^{2}

**Solution :**

We have ( 2.5p - 1.5q )^{2} - ( 1.5p - 2.5q )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} and ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here , a = 2.5p and b = 1.5q and a = 1.5p and b = 2.5q

So that,

[{ ( 2.5p )^{2} - 2 × 2.5p × 1.5q + ( 1.5q )^{2} } - { ( 1.5p )^{2} - 2 × 1.5q × 2.5p + ( 2.5q )^{2} }]

= 6.25p^{2} - 7.5pq + 2.25q^{2} - 2.25p^{2} + 7.5pq - 6.25q^{2} [ because = - 7.5pq + 7.5pq = 0 ]

= ( 6.25p^{2} - 2.25p^{2} ) - ( 2.25q^{2} - 6.25q^{2} )

= 4p^{2} - 4q^{2}

So, answer is 4p^{2} - 4q^{2}

**(vi) **( ab + bc )^{2} - 2ab^{2}c

**Solution :**

We have ( ab + bc )^{2} - 2ab^{2}c

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here , a = ab and b = bc

So that,

( ab )^{2} + 2 × ab × bc + ( bc )^{2} - 2ab^{2}c

= a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2} - 2ab^{2}c [ because = 2ab^{2}c - 2ab^{2}c = 0 ]

= a^{2}b^{2} + b^{2}c^{2}

So, answer is a^{2}b^{2} + b^{2}c^{2}

**(vii) **( m^{2} - n^{2}m )^{2} + 2m^{3}n^{2}

**Solution :**

We have ( m^{2} - n^{2}m )^{2} + 2m^{3}n^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here , a = m^{2} and b = n^{2}m

So that,

( m^{2} )^{2} - 2 × m^{2} × n^{2}m + ( n^{2}m )^{2} + 2m^{3}n^{2}

= m^{4} - 2m^{3}n^{2} + n^{4}m^{2} + 2m^{3}n^{2} [ because = - 2m^{3}n^{2} + 2m^{3}n^{2} = 0 ]

= m^{4} + n^{4}m^{2}

So, answer is m^{4} + n^{4}m^{2}

**Q5. **Show that.

**(i) **( 3x + 7 )^{2} - 84x = ( 3x - 7 )^{2}

**Solution :**

We have ( 3x + 7 )^{2} - 84x = ( 3x - 7 )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

LHS = ( 3x + 7 )^{2} - 84x = RHS = ( 3x - 7 )^{2}

= ( 3x )^{2} + 2 × 3x × 7 + ( 7 ) - 84x

= 9x^{2} + 42x + 49 - 84x [ because = 42x - 84x = - 42 ]

= 9x^{2} - 42x + 49

= ( 3x - 7 )^{2} = ( 3x - 7 )^{2}

= LHS = RHS

Hence proved

**(ii) **( 9p - 5q )^{2} + 180pq = ( 9p + 5q )^{2}

**Solution :**

We have ( 9p - 5q )^{2} + 180pq = ( 9p + 5q )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

LHS = ( 9p - 5q )^{2} + 180pq = RHS = ( 9p + 5q )^{2}

( 9p )^{2} - 2 × 9p × 5q + ( 5q )^{2} + 180pq

81p^{2} - 90pq + 25q^{2} + 180pq [ because = - 90pq + 180pq = 0 ]

= 81p^{2} + 90pq + 25q^{2}

= ( 9p + 5q )^{2} = ( 9p + 5q )^{2}

= LHS = RHS

Hence proved

**(iii) **\(\displaystyle {{\left( {\frac{4}{3}m\,-\,\frac{3}{4}n} \right)}^{2}}\,+\,2mn\,=\,\frac{{16}}{9}{{m}^{2}}\,+\,\frac{9}{{16}}{{n}^{2}}\)

**Solution :**

We have \(\displaystyle {{\left( {\frac{4}{3}m\,-\,\frac{3}{4}n} \right)}^{2}}\,+\,2mn\,=\,\frac{{16}}{9}{{m}^{2}}\,+\,\frac{9}{{16}}{{n}^{2}}\)

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

LHS = \(\displaystyle {{\left( {\frac{4}{3}m\,-\,\frac{3}{4}n} \right)}^{2}}\,+\,2mn\) = RHS = \(\displaystyle \frac{{16}}{9}{{m}^{2}}\,+\,\frac{9}{{16}}{{n}^{2}}\)

\(\displaystyle {{\left( {\frac{4}{3}m} \right)}^{2}}\,-\,2\,\times \,\frac{4}{3}m\,\times \,\frac{3}{4}n\,+\,{{\left( {\frac{3}{4}n\,} \right)}^{2}}\,+\,2mn\)

=

= \(\displaystyle \frac{{16}}{9}{{m}^{2}}\,-\,2mn\,+\,\frac{9}{{16}}{{n}^{2}}\,+\,2mn\) [ because = \(\displaystyle -\,2mn\) + \(\displaystyle \,2mn\) = 0 ]

= \(\displaystyle \frac{{16}}{9}{{m}^{2}}\,+\,\frac{9}{{16}}{{n}^{2}}\,\) = \(\displaystyle \frac{{16}}{9}{{m}^{2}}\,+\,\frac{9}{{16}}{{n}^{2}}\,\)

= LHS = RHS

Hence proved

**(iv)** ( 4pq + 3q )^{2} - ( 4pq - 3q )^{2} = 48pq^{2}

**Solution :**

We have ( 4pq + 3q )^{2} - ( 4pq - 3q )^{2} = 48pq^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} and ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

LHS = ( 4pq + 3q )^{2} - ( 4pq - 3q )^{2} = RHS = 48pq^{2}

[{ ( 4pq )^{2} + 2 × 4pq × 3q + ( 3q )^{2} } - { ( 4pq )^{2} - 2 × 4pq × 3q + ( 3q )^{2} }]

= 16p^{2}q^{2} + 24pq^{2} + 9q^{2} - 16p^{2}q^{2} + 24pq^{2} - 9q^{2} [ because = 16p^{2}q^{2} - 16p^{2}q^{2} = o and 9q^{2} - 9q^{2} = 0 ]

= 48pq^{2} = 48pq^{2}

= LHS = RHS

Hence proved

**(v) **( a - b ) ( a + b ) + ( b - c ) ( b + c ) + ( c - a ) ( c + a ) = 0

**Solution :**

We have ( a - b ) ( a + b ) + ( b - c ) ( b + c ) + ( c - a ) ( c + a ) = 0

[ we have to the multiply the first bracket by the second bracket ]

a ( a + b ) - b ( a + b ) + b ( b + c ) - c ( b + c ) + c ( c + a ) - a ( c + a ) = 0

= a^{2} + ab - ab - b^{2} + b^{2} + bc - bc - c^{2} + c^{2} + ac - ac - a^{2} = 0

= ( a^{2} - a^{2} ) ( ab - ab ) ( - b^{2} + b^{2} ) ( bc - bc ) ( - c^{2} + c^{2} ) ( ac - ac ) = 0

= ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) = 0

= 0 = 0

= LHS = RHS

Hence proved

**Q6. **Using identities evaluate.

**(i) **71^{2}

**Solution :**

We have 71^{2} = ( 70 + 1 )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here, a = 70 and b = 1

So that,

( 70 )^{2} + 2 × 70 × 1 + ( 1 )^{2}

= 4900 + 140 + 1

= 5041

So, answer is 5041

**(ii) **99^{2}

**Solution :**

We have 99^{2} = ( 100 - 1 )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here, a = 100 and b = 1

So that,

( 100 )^{2} - 2 × 100 × 1 + ( 1 )^{2}

= 10000 - 200 + 1

= 9800 + 1

= 9801

So, answer is 9801

**(iii) **102^{2}

**Solution :**

We have 102^{2} = ( 100 + 2 )^{2}

[ using identity = ( a + b )^{2} = a^{2} + 2ab + b^{2} ]

Here, a = 100 and b = 2

So that,

( 100 )^{2} + 2 × 100 × 2 + ( 2 )^{2}

= 10000 + 400 + 4

= 10404

So, answer is 10404

**(iv) **998^{2}

**Solution :**

We have 998^{2} = ( 1000 - 2 )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here, a = 1000 and b = 2

So that,

( 1000 )^{2} - 2 × 1000 × 2 + ( 2 )^{2}

= 1000000 - 4000 + 4

= 996000 + 4

= 996004

So, answer is 996004

**(v)** 5.2^{2}

We have 5.2^{2} = ( 5 + 0.2 )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here, a = 5 and b = 0.2

So that,

( 5 )^{2} + 2 × 5 × 0.2 + ( 0.2 )^{2}

= 25 + 2 + 0.04

= 27.04

So, answer is 27.04

**(vi) **297 × 303

**Solution :**

We have 297 × 303 = ( 300 - 3 ) ( 300 + 3 )

[ using identity = ( a - b ) ( a + b ) = a^{2} - b^{2} ]

Here, a = 300 and b = 3

So that,

( 300 )^{2} - ( 3 )^{2}

= 90000 - 9

= 89991

So, answer is 89991

**(vii) **78 × 82

**Solution :**

We have 78 × 82 = ( 80 - 2 ) ( 80 + 2 )

[ using identity = ( a - b ) ( a + b ) = a^{2} - b^{2} ]

Here, a = 80 and b = 2

So that,

( 80 )^{2} - ( 2 )^{2}

= 6400 - 4

= 6396

So, answer is 6396

**(viii) **8.9^{2}

**Solution :**

We have 8.9^{2} = ( 9 - 0.1 )^{2}

[ using identity = ( a - b )^{2} = a^{2} - 2ab + b^{2} ]

Here, a = 9 and b = 0.1

So that,

( 9 )^{2} - 2 × 9 × 0.1 + ( 0.1 )

= 81 - 1.8 + 0.01

= 79.2 + 0.01

= 79.21

So, answer is 79.21

**(ix) **10.5 × 9.5

**Solution :**

We have 10.5 × 9.5 = ( 10 + 0.5 ) ( 10 - 0.5 )

[ using identity = ( a - b ) ( a + b ) = a^{2} - b^{2} ]

Here, a = 10 and b = 0.5

So that,

( 10 )^{2} - ( 0.5 )^{2}

= 100 - 0.25

= 99.75

So, answer is 99. 75

**Q7. **Using a^{2} - b^{2} = ( a + b ) ( a - b ), find

**(i)** 5l^{2} - 49^{2}

**Solution :**

We have 51^{2} - 49^{2}

[ using identity = a^{2} - b^{2} = ( a + b ) ( a - b ) ]

Here, a = 51 and b = 49

So that,

( 51 + 49 ) ( 51 - 49 )

= ( 100 ) ( 2 )

= 100 × 2

= 200

So, answer is 200

**(ii)** ( 1.02 )^{2} - ( 0.98 )^{2}

**Solution :**

We have ( 1.02 )^{2} - ( 0.98 )^{2}

[ using identity = a^{2} - b^{2} = ( a + b ) ( a - b ) ]

Here, a = 1.02 and b = 0.98

So that,

( 1.02 + 0.98 ) ( 1.02 - 0.98 )

= ( 2 ) ( 0.04 )

= 2 × 0.04

= 0.08

So, answer is 0.08

**(iii) **153^{2} - 147^{2}

**Solution :**

We have 153^{2} - 147^{2}

[ using identity = a^{2} - b^{2} = ( a + b ) ( a - b ) ]

Here, a = 153 and b = 147

So that,

( 153 + 147 ) ( 153 - 147 )

= ( 300 ) ( 6 )

= 300 × 6

= 1800

So, answer is 1800

**(iv) **12.1^{2} - 7.9^{2}

**Solution :**

We have 12.1^{2} - 7.9^{2}

[ using identity = a^{2} - b^{2} = ( a + b ) ( a - b ) ]

Here, a = 12.1 and b = 7.9

So that,

( 12.1 + 7.9 ) ( 12.1 - 7.9 )

= ( 20 ) ( 4.2 )

= 20 × 4.2

= 84

So, answer is 84

**Q8.** Using ( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab, find

**(i) **103 × 104

**Solution :**

We have 103 × 104

( 100 + 3 ) ( 100 + 4 )

[ using identity = ( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

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

So that,

( 100 )^{2} + ( 3 + 4 ) 100 + 3 × 4

= 10000 + ( 7 ) 100 + 12

= 10000 + 7 × 100 + 12

= 10000 + 700 + 12

= 10712

So, answer is 10712

**(ii) **5.1 × 5.2

**Solution :**

We have 5.1 × 5.2

( 5 + 0.1 ) ( 5 + 0.2 )

[ using identity = ( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

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

So that,

( 5 )^{2} + ( 0.1 + 0.2 ) 5 + 0.1 × 0.2

= 25 + ( 0.3 ) 5 + 0.02

= 25 + 1.5 + 0.02

= 26.57

So, answer is 26.57

**(iii) **103 × 98

**Solution :**

We have 103 × 98

( 100 + 3 ) ( 100 - 2 )

[ using identity = ( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

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

So that,

( 100 )^{2} + ( 3 - 2 )100 + 3 × ( - 2 )

= 10000 + 1 × 100 - 6

= 10000 + 100 - 6

= 10100 - 6

= 10094

So, answer is 10094

**(iv)** 9.7 × 9.8

**Solution :**

We have 9.7 × 9.8

( 10 - 0.3 ) ( 10 - 0.2 )

[ using identity = ( x + a ) ( x + b ) = x^{2} + ( a + b ) x + ab ]

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

So that,

( 10 )^{2} + ( - 0.3 - 0.2 )10 + ( - 0.3 ) × ( - 0.2 )

= 100 - 0.5 × 10 + 0.06

= 100 - 5 + 0.06

= 100.06 - 5

= 95.06

So, answer is 95.06 .

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities :

- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.1
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.2
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.3
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.4
- NCERT Class 8 Maths Algebraic Expressions and Identities Exercise 9.5