# NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities (EX 9.5) Exercise 9.5

## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities (EX 9.5) Exercise 9.5

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

Exercise 9.5

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

i. $\left( {x + 3} \right)\left( {x + 3} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {x + 3} \right)\left( {x + 3} \right) = {\left( {x + 3} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {x + 3} \right)^2} = {\left( x \right)^2} + 2\left( x \right)\left( 3 \right) + {\left( 3 \right)^2}$

${\left( {x + 3} \right)^2} = {x^2} + 6x + 9$

ii. $\left( {2y + 5} \right)\left( {2y + 5} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {2y + 5} \right)\left( {2y + 5} \right) = {\left( {2y + 5} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {2y + 5} \right)^2} = {\left( {2y} \right)^2} + 2\left( {2y} \right)\left( 5 \right) + {\left( 5 \right)^2}$

${\left( {2y + 5} \right)^2} = 4{y^2} + 20y + 25$

iii. $\left( {2a - 7} \right)\left( {2a - 7} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {2a - 7} \right)\left( {2a - 7} \right) = {\left( {2a - 7} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {2a - 7} \right)^2} = {\left( {2a} \right)^2} - 2\left( {2a} \right)\left( 7 \right) + {\left( 7 \right)^2}$

${\left( {2a - 7} \right)^2} = 4{a^2} - 28a + 49$

iv. $\left( {3a - \dfrac{1}{2}} \right)\left( {3a - \dfrac{1}{2}} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {3a - \dfrac{1}{2}} \right)\left( {3a - \dfrac{1}{2}} \right) = {\left( {3a - \dfrac{1}{2}} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {3a - \dfrac{1}{2}} \right)^2} = {\left( {3a} \right)^2} - 2\left( {3a} \right)\left( {\dfrac{1}{2}} \right) + {\left( {\dfrac{1}{2}} \right)^2}$

${\left( {3a - \dfrac{1}{2}} \right)^2} = 9{a^2} - 3a + \dfrac{1}{4}$

v. $\left( {1.1m - 0.4} \right)\left( {1.1m + 0.4} \right)$

Ans: On writing the given expression using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$, we get,$\left( {1.1m - 0.4} \right)\left( {1.1m + 0.4} \right) = {\left( {1.1m} \right)^2} - {\left( {0.4} \right)^2}$

On simplifying the above equation further, we get,

$\left( {1.1m - 0.4} \right)\left( {1.1m + 0.4} \right) = 1.21{m^2} - 0.16$

vi. $\left( {{a^2} + {b^2}} \right)\left( { - {a^2} + {b^2}} \right)$

Ans: Using the commutative rule and writing the given expression using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$, we get,

$\left( {{a^2} + {b^2}} \right)\left( { - {a^2} + {b^2}} \right) = \left( {{b^2} + {a^2}} \right)\left( {{b^2} - {a^2}} \right)$

$\left( {{a^2} + {b^2}} \right)\left( { - {a^2} + {b^2}} \right) = {\left( {{b^2}} \right)^2} - {\left( {{a^2}} \right)^2}$

On simplifying the above equation further, we get,

$\left( {{a^2} + {b^2}} \right)\left( { - {a^2} + {b^2}} \right) = {b^4} - {a^4}\,$

vii. $\left( {6x - 7} \right)\left( {6x + 7} \right)$

Ans: On writing the given expression using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$, we get,

$\left( {6x - 7} \right)\left( {6x + 7} \right) = {\left( {6x} \right)^2} - {\left( 7 \right)^2}$

On simplifying the above equation further, we get,

$\left( {6x - 7} \right)\left( {6x + 7} \right) = 36{x^2} - 49$

viii. $\left( { - a + c} \right)\left( { - a + c} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( { - a + c} \right)\left( { - a + c} \right) = {\left( { - a + c} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

$\left( { - a + c} \right)\left( { - a + c} \right) = {\left( { - a} \right)^2} + 2\left( { - a} \right)\left( c \right) + {\left( c \right)^2}$

$\left( { - a + c} \right)\left( { - a + c} \right) = {a^2} - 2ac + {c^2}$

ix. $\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right) = {\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

$\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right) = {\left( {\dfrac{x}{2}} \right)^2} + 2\left( {\dfrac{x}{2}} \right)\left( {\dfrac{{3y}}{4}} \right) + {\left( {\dfrac{{3y}}{4}} \right)^2}$

$\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right)\left( {\dfrac{x}{2} + \dfrac{{3y}}{4}} \right) = \dfrac{{{x^2}}}{4} + \dfrac{{3xy}}{4} + \dfrac{{9{y^2}}}{{16}}$

x. $\left( {7a - 9b} \right)\left( {7a - 9b} \right)$

Ans: On writing the given expression in the form of a suitable identity, we get,

$\left( {7a - 9b} \right)\left( {7a - 9b} \right) = {\left( {7a - 9b} \right)^2}$

On simplifying the above equation using the identity ${\left( {a + b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

$\left( {7a - 9b} \right)\left( {7a - 9b} \right) = {\left( {7a} \right)^2} - 2\left( {7a} \right)\left( {9b} \right) + {\left( {9b} \right)^2}$

$\left( {7a - 9b} \right)\left( {7a - 9b} \right) = 49{a^2} - 126ab + 81{b^2}$

2. Use the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$to find the following products.

(i). $\left( {x + 3} \right)\left( {x + 7} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {x + 3} \right)\left( {x + 7} \right) = {x^2} + \left( {3 + 7} \right)x + \left( 3 \right)\left( 7 \right)$

On simplifying the above equation further, we get,

$\left( {x + 3} \right)\left( {x + 7} \right) = {x^2} + 10x + 21$

(ii). $\left( {4x + 5} \right)\left( {4x + 1} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {4x + 5} \right)\left( {4x + 1} \right) = {\left( {4x} \right)^2} + \left( {5 + 1} \right)\left( {4x} \right) + \left( 5 \right)\left( 1 \right)$

On simplifying the above equation further, we get,

$\left( {4x + 5} \right)\left( {4x + 1} \right) = 16{x^2} + 6\left( {4x} \right) + 5$

$\left( {4x + 5} \right)\left( {4x + 1} \right) = 16{x^2} + 24x + 5$

(iii). $\left( {4x - 5} \right)\left( {4x - 1} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {4x - 5} \right)\left( {4x - 1} \right) = {\left( {4x} \right)^2} + \left[ {\left( { - 5} \right) + \left( { - 1} \right)} \right]\left( {4x} \right) + \left( { - 5} \right)\left( { - 1} \right)$

On simplifying the above equation further, we get,

$\left( {4x - 5} \right)\left( {4x - 1} \right) = 16{x^2} + \left( { - 5 - 1} \right)\left( {4x} \right) + 5$

$\left( {4x - 5} \right)\left( {4x - 1} \right) = 16{x^2} + \left( { - 6} \right)\left( {4x} \right) + 5$

$\left( {4x - 5} \right)\left( {4x - 1} \right) = 16{x^2} - 24x + 5$

(iv). $\left( {4x + 5} \right)\left( {4x - 1} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {4x + 5} \right)\left( {4x - 1} \right) = {\left( {4x} \right)^2} + \left[ {\left( 5 \right) + \left( { - 1} \right)} \right]\left( {4x} \right) + \left( 5 \right)\left( { - 1} \right)$

On simplifying the above equation further, we get,

$\left( {4x + 5} \right)\left( {4x - 1} \right) = 16{x^2} + \left( {5 - 1} \right)\left( {4x} \right) - 5$

$\left( {4x + 5} \right)\left( {4x - 1} \right) = 16{x^2} + \left( 4 \right)\left( {4x} \right) - 5$

$\left( {4x + 5} \right)\left( {4x - 1} \right) = 16{x^2} + 16x - 5$

(v). $\left( {2x + 5y} \right)\left( {2x + 3y} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {2x + 5y} \right)\left( {2x + 3y} \right) = {\left( {2x} \right)^2} + \left( {5y + 3y} \right)\left( {2x} \right) + \left( {5y} \right)\left( {3y} \right)$

On simplifying the above equation further, we get,

$\left( {2x + 5y} \right)\left( {2x + 3y} \right) = 4{x^2} + \left( {8y} \right)\left( {2x} \right) + 15{y^2}$

$\left( {2x + 5y} \right)\left( {2x + 3y} \right) = 4{x^2} + 16xy + 15{y^2}$

(vi). $\left( {2{a^2} + 9} \right)\left( {2{a^2} + 5} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {2{a^2} + 9} \right)\left( {2{a^2} + 5} \right) = {\left( {2{a^2}} \right)^2} + \left( {9 + 5} \right)\left( {2{a^2}} \right) + \left( 9 \right)\left( 5 \right)$

On simplifying the above equation further, we get,

$\left( {2{a^2} + 9} \right)\left( {2{a^2} + 5} \right) = 4{a^4} + \left( {14} \right)\left( {2{a^2}} \right) + 45$

$\left( {2{a^2} + 9} \right)\left( {2{a^2} + 5} \right) = 4{a^4} + 28{a^2} + 45$

(vii). $\left( {xyz - 4} \right)\left( {xyz - 2} \right)$

Ans: On writing the given expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {xyz - 4} \right)\left( {xyz - 2} \right) = {\left( {xyz} \right)^2} + \left[ {\left( { - 4} \right) + \left( { - 2} \right)} \right]\left( {xyz} \right) + \left( { - 4} \right)\left( { - 2} \right)$

On simplifying the above equation further, we get,

$\left( {xyz - 4} \right)\left( {xyz - 2} \right) = {x^2}{y^2}{z^2} + \left( { - 4 - 2} \right)\left( {xyz} \right) + 8$

$\left( {xyz - 4} \right)\left( {xyz - 2} \right) = {x^2}{y^2}{z^2} - 6xyz + 8$

3. Find the following squares by using the identities.

(i). ${\left( {b - 7} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {b - 7} \right)^2} = {\left( b \right)^2} - 2\left( b \right)\left( 7 \right) + {\left( 7 \right)^2}$

${\left( {b - 7} \right)^2} = {b^2} - 14b + 49$

(ii). ${\left( {xy + 3z} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {xy + 3z} \right)^2} = {\left( {xy} \right)^2} + 2\left( {xy} \right)\left( {3z} \right) + {\left( {3z} \right)^2}$

${\left( {xy + 3z} \right)^2} = {x^2}{y^2} + 6xyz + 9{z^2}$

(iii). ${\left( {6{x^2} - 5y} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {6{x^2} - 5y} \right)^2} = {\left( {6{x^2}} \right)^2} - 2\left( {6{x^2}} \right)\left( {5y} \right) + {\left( {5y} \right)^2}$

${\left( {6{x^2} - 5y} \right)^2} = 36{x^4} - 2\left( {30{x^2}y} \right) + 25{y^2}$

${\left( {6{x^2} - 5y} \right)^2} = 36{x^4} - 60{x^2}y + 25{y^2}$

(iv). ${\left( {\dfrac{2}{3}m + \dfrac{3}{2}n} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {\dfrac{2}{3}m + \dfrac{3}{2}n} \right)^2} = {\left( {\dfrac{2}{3}m} \right)^2} + 2\left( {\dfrac{2}{3}m} \right)\left( {\dfrac{3}{2}n} \right) + {\left( {\dfrac{3}{2}n} \right)^2}$

${\left( {\dfrac{2}{3}m + \dfrac{3}{2}n} \right)^2} = \dfrac{4}{9}{m^2} + 2mn + \dfrac{9}{4}{n^2}$

(v). ${\left( {0.4p - 0.5q} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {0.4p - 0.5q} \right)^2} = {\left( {0.4p} \right)^2} - 2\left( {0.4p} \right)\left( {0.5q} \right) + {\left( {0.5q} \right)^2}$

${\left( {0.4p - 0.5q} \right)^2} = 0.16{p^2} - 2\left( {0.2pq} \right) + 0.25{q^2}$

${\left( {0.4p - 0.5q} \right)^2} = 0.16{p^2} - 0.4pq + 0.25{q^2}$

(vi). ${\left( {2xy + 5y} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {2xy + 5y} \right)^2} = {\left( {2xy} \right)^2} + 2\left( {2xy} \right)\left( {5y} \right) + {\left( {5y} \right)^2}$

${\left( {2xy + 5y} \right)^2} = 4{x^2}{y^2} + 2\left( {10x{y^2}} \right) + 25{y^2}$

${\left( {2xy + 5y} \right)^2} = 4{x^2}{y^2} + 20x{y^2} + 25{y^2}$

4. Simplify the following expressions.

(i). ${\left( {{a^2} - {b^2}} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {{a^2} - {b^2}} \right)^2} = {\left( {{a^2}} \right)^2} - 2\left( {{a^2}} \right)\left( {{b^2}} \right) + {\left( {{b^2}} \right)^2}$

${\left( {{a^2} - {b^2}} \right)^2} = {a^4} - 2{a^2}{b^2} + {b^4}$

(ii). ${\left( {2x + 5} \right)^2} - {\left( {2x - 5} \right)^2}$

Ans: On expanding the term ${\left( {2x + 5} \right)^2}$ using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and expanding the term ${\left( {2x - 5} \right)^2}$ using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {2x + 5} \right)^2} - {\left( {2x - 5} \right)^2} = {\left( {2x} \right)^2} + 2\left( {2x} \right)\left( 5 \right) + {\left( 5 \right)^2} - \left[ {{{\left( {2x} \right)}^2} - 2\left( {2x} \right)\left( 5 \right) + {{\left( 5 \right)}^2}} \right]$

${\left( {2x + 5} \right)^2} - {\left( {2x - 5} \right)^2} = 4{x^2} + 20x + 25 - \left[ {4{x^2} - 20x + 25} \right]$

${\left( {2x + 5} \right)^2} - {\left( {2x - 5} \right)^2} = 4{x^2} + 20x + 25 - 4{x^2} + 20x - 25$

${\left( {2x + 5} \right)^2} - {\left( {2x - 5} \right)^2} = 40x$

(iii). ${\left( {7m - 8n} \right)^2} + {\left( {7m + 8n} \right)^2}$

Ans: On expanding the term ${\left( {7m - 8n} \right)^2}$ using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and expanding the term ${\left( {7m + 8n} \right)^2}$ using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {7m - 8n} \right)^2} + {\left( {7m + 8n} \right)^2} = {\left( {7m} \right)^2} - 2\left( {7m} \right)\left( {8n} \right) + {\left( {8n} \right)^2} + \left[ {{{\left( {7m} \right)}^2} + 2\left( {7m} \right)\left( {8n} \right) + {{\left( {8n} \right)}^2}} \right]$

${\left( {7m - 8n} \right)^2} + {\left( {7m + 8n} \right)^2} = 49{m^2} - 112mn + 64{n^2} + \left[ {49{m^2} + 112mn + 64{n^2}} \right]$${\left( {7m - 8n} \right)^2} + {\left( {7m + 8n} \right)^2} = 49{m^2} - 112mn + 64{n^2} + 49{m^2} + 112mn + 64{n^2}$

${\left( {7m - 8n} \right)^2} + {\left( {7m + 8n} \right)^2} = 98{m^2} + 128{n^2}$

(iv). ${\left( {4m + 5n} \right)^2} + {\left( {5m + 4n} \right)^2}$

Ans: On expanding the terms ${\left( {4m + 5n} \right)^2}$ and ${\left( {5m + 4n} \right)^2}$ using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {4m + 5n} \right)^2} + {\left( {5m + 4n} \right)^2} = {\left( {4m} \right)^2} + 2\left( {4m} \right)\left( {5n} \right) + {\left( {5n} \right)^2} + \left[ {{{\left( {5m} \right)}^2} + 2\left( {5m} \right)\left( {4n} \right) + {{\left( {4n} \right)}^2}} \right]$

${\left( {4m + 5n} \right)^2} + {\left( {5m + 4n} \right)^2} = 16{m^2} + 40mn + 25{n^2} + \left[ {25{m^2} + 40mn + 16{n^2}} \right]$

${\left( {4m + 5n} \right)^2} + {\left( {5m + 4n} \right)^2} = 16{m^2} + 40mn + 25{n^2} + 25{m^2} + 40mn + 16{n^2}$${\left( {4m + 5n} \right)^2} + {\left( {5m + 4n} \right)^2} = 41{m^2} + 80mn + 41{n^2}$

(v). ${\left( {2.5p - 1.5q} \right)^2} -{\left( {1.5p - 2.5q} \right)^2}$

Ans: On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

Solving ${\left( {2.5p - 1.5q} \right)^2}$ we get

$=\, 2.5p^2+1.5q^2-2(2.5p)(1.5q)$

$={\dfrac{25}{10}}^2\times p^2+{\dfrac{15}{10}}^2\times q^2 -(2\times\dfrac{25}{10}\times\dfrac{15}{10})\times(p \times q)$

$=\dfrac{625}{100}p^2+\dfrac{225}{100} q^2 - \dfrac{750}{100} pq$

Solving ${\left( {1.5p - 2.5q} \right)^2}$ we get

$=\, 1.5p^2\,+\,2.5q^2\,-\,2(1.5p)(2.5q)$

$={\dfrac{15}{10}}^2\times p^2+{\dfrac{25}{10}}^2\times q^2 -(2\times\dfrac{15}{10}\times\dfrac{25}{10})\times(p \times q)$

$=\dfrac{225}{100}p^2+\dfrac{625}{100} q^2 - \dfrac{750}{100} pq$

$\therefore$  solving the given equation we get,

${\left( {2.5p - 1.5q} \right)^2} -{\left( {1.5p - 2.5q} \right)^2}$ = $\dfrac{625}{100}p^2+\dfrac{225}{100} q^2 - \dfrac{750}{100} pq -$  ($\dfrac{225}{100}p^2+\dfrac{625}{100} q^2 - \dfrac{750}{100} pq$)

$=\dfrac{400}{100} p^2 - \dfrac{400}{100} q^2$

$=4p^2-4q^2$

(vi). ${\left( {ab + bc} \right)^2} - 2a{b^2}c$

Ans: On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {ab + bc} \right)^2} - 2a{b^2}c = {\left( {ab} \right)^2} + 2\left( {ab} \right)\left( {bc} \right) + {\left( {bc} \right)^2} - 2a{b^2}c$

${\left( {ab + bc} \right)^2} - 2a{b^2}c = {a^2}{b^2} + 2a{b^2}c + {b^2}{c^2} - 2a{b^2}c$

${\left( {ab + bc} \right)^2} - 2a{b^2}c = {a^2}{b^2} + {b^2}{c^2}$

(vii). ${\left( {{m^2} - {n^2}m} \right)^2} + 2{m^3}{n^2}$

Ans: On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {{m^2} - {n^2}m} \right)^2} + 2{m^3}{n^2} = {\left( {{m^2}} \right)^2} - 2\left( {{m^2}} \right)\left( {{n^2}m} \right) + {\left( {{n^2}m} \right)^2} + 2{m^3}{n^2}$

${\left( {{m^2} - {n^2}m} \right)^2} + 2{m^3}{n^2} = {m^4} - 2{m^3}{n^2} + {n^4}{m^2} + 2{m^3}{n^2}$

${\left( {{m^2} - {n^2}m} \right)^2} + 2{m^3}{n^2} = {m^4} + {n^4}{m^2}$

5. Show that the following equations are true.

(i). ${\left( {3x + 7} \right)^2} - 84x = {\left( {3x - 7} \right)^2}$

Ans: On considering the ${\text{LHS}}$ of the expression, we get,

${\text{LHS}} = {\left( {3x + 7} \right)^2} - 84x$

On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\text{LHS}} = {\left( {3x + 7} \right)^2} - 84x$

${\text{LHS}} = {\left( {3x} \right)^2} + 2\left( {3x} \right)\left( 7 \right) + {\left( 7 \right)^2} - 84x$

${\text{LHS}} = 9{x^2} + 42x + 49 - 84x$

${\text{LHS}} = 9{x^2} - 42x + 49$

On considering the ${\text{RHS}}$ of the expression, we get,

${\text{RHS}} = {\left( {3x - 7} \right)^2}$

On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\text{RHS}} = {\left( {3x - 7} \right)^2}$

${\text{RHS}} = {\left( {3x} \right)^2} - 2\left( {3x} \right)\left( 7 \right) + {\left( 7 \right)^2}$

${\text{RHS}} = 9{x^2} - 42x + 49$

As ${\text{LHS = RHS}}$, so the given expression is proved.

(ii). ${\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}$

Ans: On considering the ${\text{LHS}}$ of the expression, we get,

${\text{LHS}} = {\left( {9p - 5q} \right)^2} + 180pq$

On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\text{LHS}} = {\left( {9p - 5q} \right)^2} + 180pq$

${\text{LHS}} = {\left( {9p} \right)^2} - 2\left( {9p} \right)\left( {5q} \right) + {\left( {5q} \right)^2} + 180pq$

${\text{LHS}} = 81{p^2} - 90pq + 25{q^2} + 180pq$

${\text{LHS}} = 81{p^2} + 90pq + 25{q^2}$

On considering the ${\text{RHS}}$ of the expression, we get,

${\text{RHS}} = {\left( {9p + 5q} \right)^2}$

On expanding the given expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\text{RHS}} = {\left( {9p + 5q} \right)^2}$

${\text{RHS}} = {\left( {9p} \right)^2} + 2\left( {9p} \right)\left( {5q} \right) + {\left( {5q} \right)^2}$

${\text{RHS}} = 81{p^2} + 90pq + 25{q^2}$

As ${\text{LHS = RHS}}$, so the given expression is proved.

(iii). ${\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2}$

Ans: On considering the ${\text{LHS}}$ of the expression, we get,

${\text{LHS}} = {\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn$

On expanding the given expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\text{LHS}} = {\left( {\dfrac{4}{3}m - \dfrac{3}{4}n} \right)^2} + 2mn$

${\text{LHS}} = {\left( {\dfrac{4}{3}m} \right)^2} - 2\left( {\dfrac{4}{3}m} \right)\left( {\dfrac{3}{4}n} \right) + {\left( {\dfrac{3}{4}n} \right)^2} + 2mn$

${\text{LHS}} = \dfrac{{16}}{9}{m^2} - 2mn + \dfrac{9}{{16}}{n^2} + 2mn$

${\text{LHS}} = \dfrac{{16}}{9}{m^2} + \dfrac{9}{{16}}{n^2}$

${\text{LHS}} = {\text{RHS}}$

As ${\text{LHS = RHS}}$, so the given expression is proved.

(iv). ${\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2} = 48p{q^2}$

Ans: On considering the ${\text{LHS}}$ of the expression, we get,

${\text{LHS}} = {\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2}$

On expanding the term ${\left( {4pq + 3q} \right)^2}$ using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and expanding the term ${\left( {4pq - 3q} \right)^2}$ using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\text{LHS}} = {\left( {4pq + 3q} \right)^2} - {\left( {4pq - 3q} \right)^2}$

${\text{LHS}} = {\left( {4pq} \right)^2} + 2\left( {4pq} \right)\left( {3q} \right) + {\left( {3q} \right)^2} - \left[ {{{\left( {4pq} \right)}^2} - 2\left( {4pq} \right)\left( {3q} \right) + {{\left( {3q} \right)}^2}} \right]$

${\text{LHS}} = 16{p^2}{q^2} + 24p{q^2} + 9{q^2} - \left[ {16{p^2}{q^2} - 24p{q^2} + 9{q^2}} \right]$

${\text{LHS}} = 16{p^2}{q^2} + 24p{q^2} + 9{q^2} - 16{p^2}{q^2} + 24p{q^2} - 9{q^2}$

${\text{LHS}} = 48p{q^2}$

${\text{LHS}} = {\text{RHS}}$

As ${\text{LHS = RHS}}$, so the given expression is proved.

(v). $\left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right) = 0$

Ans: On considering the ${\text{LHS}}$ of the expression, we get,

${\text{LHS}} = \left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right)$

On simplifying the terms using the identity $\left( {x - y} \right)\left( {x + y} \right) = {x^2} - {y^2}$ and solving it further, we get,

${\text{LHS}} = \left( {a - b} \right)\left( {a + b} \right) + \left( {b - c} \right)\left( {b + c} \right) + \left( {c - a} \right)\left( {c + a} \right)$

${\text{LHS}} = \left( {{a^2} - {b^2}} \right) + \left( {{b^2} - {c^2}} \right) + \left( {{c^2} - {a^2}} \right)$

${\text{LHS}} = \left( {{a^2} - {a^2}} \right) + \left( {{b^2} - {b^2}} \right) + \left( {{c^2} - {c^2}} \right)$

${\text{LHS}} = 0$

${\text{LHS}} = {\text{RHS}}$

As ${\text{LHS = RHS}}$, so the given expression is proved.

6. Using the different algebraic identities, evaluate the following.

(i). ${71^2}$

Ans: On expressing ${71^2}$ as the square of the sum of two numbers, we get,

${71^2} = {\left( {70 + 1} \right)^2}$

On expanding the above expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {70 + 1} \right)^2} = {\left( {70} \right)^2} + 2\left( {70} \right)\left( 1 \right) + {\left( 1 \right)^2}$

${\left( {70 + 1} \right)^2} = 4900 + 140 + 1$

${\left( {70 + 1} \right)^2} = 5041$

(ii). ${99^2}$

Ans: On expressing ${99^2}$ as the square of the difference of two numbers, we get,

${99^2} = {\left( {100 - 1} \right)^2}$

On expanding the above expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {100 - 1} \right)^2} = {\left( {100} \right)^2} - 2\left( {100} \right)\left( 1 \right) + {\left( 1 \right)^2}$

${\left( {100 - 1} \right)^2} = 10000 - 200 + 1$

${\left( {100 - 1} \right)^2} = 9801$

(iii). ${102^2}$

Ans: On expressing ${102^2}$ as the square of the sum of two numbers, we get,

${102^2} = {\left( {100 + 2} \right)^2}$

On expanding the above expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {100 + 2} \right)^2} = {\left( {100} \right)^2} + 2\left( {100} \right)\left( 2 \right) + {\left( 2 \right)^2}$

${\left( {100 + 2} \right)^2} = 10000 + 400 + 4$

${\left( {100 + 2} \right)^2} = 10404$

(iv). ${998^2}$

Ans: On expressing ${998^2}$ as the square of the difference of two numbers, we get,

${998^2} = {\left( {1000 - 2} \right)^2}$

On expanding the above expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {1000 - 2} \right)^2} = {\left( {1000} \right)^2} - 2\left( {1000} \right)\left( 2 \right) + {\left( 2 \right)^2}$

${\left( {1000 - 2} \right)^2} = 1000000 - 4000 + 4$

${\left( {1000 - 2} \right)^2} = 996004$

(v). ${\left( {5.2} \right)^2}$

Ans: On expressing ${\left( {5.2} \right)^2}$ as the square of the sum of two numbers, we get,

${\left( {5.2} \right)^2} = {\left( {5.0 + 0.2} \right)^2}$

On expanding the above expression using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and solving it further, we get,

${\left( {5.0 + 0.2} \right)^2} = {\left( {5.0} \right)^2} + 2\left( {5.0} \right)\left( {0.2} \right) + {\left( {0.2} \right)^2}$

${\left( {5.0 + 0.2} \right)^2} = 25 + 2 + 0.04$

${\left( {5.0 + 0.2} \right)^2} = 27.04$

(vi). $297 \times 303$

Ans: On expressing 297 and 303 as the difference and sum of the same two numbers respectively, we get,

$297 \times 303 = \left( {300 - 3} \right) \times \left( {300 + 3} \right)$

On writing the above expression using the identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ and solving it further, we get,

$\left( {300 - 3} \right) \times \left( {300 + 3} \right) = {\left( {300} \right)^2} - {\left( 3 \right)^2}$

$\left( {300 - 3} \right) \times \left( {300 + 3} \right) = 90000 - 9$

$\left( {300 - 3} \right) \times \left( {300 + 3} \right) = 89991$

(vii). $78 \times 82$

Ans: On expressing 78 and 82 as the difference and sum of the same two numbers respectively, we get,

$78 \times 82 = \left( {80 - 2} \right) \times \left( {80 + 2} \right)$

On writing the above expression using the identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ and solving it further, we get,

$\left( {80 - 2} \right) \times \left( {80 + 2} \right) = {\left( {80} \right)^2} - {\left( 2 \right)^2}$

$\left( {80 - 2} \right) \times \left( {80 + 2} \right) = 6400 - 4$

$\left( {80 - 2} \right) \times \left( {80 + 2} \right) = 6396$

(viii). ${\left( {8.9} \right)^2}$

Ans: On expressing ${\left( {8.9} \right)^2}$ as the square of the difference of two numbers, we get,

$\,{\left( {8.9} \right)^2} = {\left( {9.0 - 0.1} \right)^2}$

On expanding the above expression using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and solving it further, we get,

${\left( {9.0 - 0.1} \right)^2} = {\left( {9.0} \right)^2} - 2\left( {9.0} \right)\left( {0.1} \right) + {\left( {0.1} \right)^2}$

${\left( {9.0 - 0.1} \right)^2} = 81 - 1.8 + 0.01$

${\left( {9.0 - 0.1} \right)^2} = 79.21$

(ix). ${\text{1}}{\text{.05}} \times {\text{9}}{\text{.5}}$

Ans: On expressing ${\text{9}}{\text{.5}}$ as the product of two numbers, we get,

$1.05 \times 9.5 = 1.05 \times 0.95 \times 10$

On expressing $1.05$ and $0.95$ as the sum and difference of the same two numbers respectively, we get,

$1.05 \times 0.95 \times 10 = \left( {1 + 0.05} \right) \times \left( {1 - 0.05} \right) \times 10$

On writing the above expression using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ and solving it further, we get,

$\left( {1 + 0.05} \right) \times \left( {1 - 0.05} \right) \times 10 = \left[ {{{\left( 1 \right)}^2} - {{\left( {0.05} \right)}^2}} \right] \times 10$

$\left( {1 + 0.05} \right) \times \left( {1 - 0.05} \right) \times 10 = \left[ {1 - 0.0025} \right] \times 10$

$\left( {1 + 0.05} \right) \times \left( {1 - 0.05} \right) \times 10 = 0.9975 \times 10$

$\left( {1 + 0.05} \right) \times \left( {1 - 0.05} \right) \times 10 = 9.975$

7. Use the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$to evaluate the following expressions.

(i). ${51^2} - {49^2}$

Ans: On writing the given expression in the form of the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we get,

${51^2} - {49^2} = \left( {51 + 49} \right)\left( {51 - 49} \right)$

On simplifying the above equation further, we get,

${51^2} - {49^2} = \left( {100} \right)\left( 2 \right)$

${51^2} - {49^2} = 200$

(ii). ${\left( {1.02} \right)^2} - {\left( {0.98} \right)^2}$

Ans: On writing the given expression in the form of the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we get,

${\left( {1.02} \right)^2} - {\left( {0.98} \right)^2} = \left( {1.02 + 0.98} \right)\left( {1.02 - 0.98} \right)$

On simplifying the above equation further, we get,

${\left( {1.02} \right)^2} - {\left( {0.98} \right)^2} = \left( 2 \right)\left( {0.04} \right)$

${\left( {1.02} \right)^2} - {\left( {0.98} \right)^2} = 0.08$

(iii). ${153^2} - {147^2}$

Ans: On writing the given expression in the form of the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we get,

${153^2} - {147^2} = \left( {153 + 147} \right)\left( {153 - 147} \right)$

On simplifying the above equation further, we get,

${153^2} - {147^2} = \left( {300} \right)\left( 6 \right)$

${153^2} - {147^2} = 1800$

(iv). ${\left( {12.1} \right)^2} - {\left( {7.9} \right)^2}$

Ans: On writing the given expression in the form of the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we get,

${\left( {12.1} \right)^2} - {\left( {7.9} \right)^2} = \left( {12.1 + 7.9} \right)\left( {12.1 - 7.9} \right)$

On simplifying the above equation further, we get,

${\left( {12.1} \right)^2} - {\left( {7.9} \right)^2} = \left( {20.0} \right)\left( {4.2} \right)$

${\left( {12.1} \right)^2} - {\left( {7.9} \right)^2} = 84$

8. Use the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$to find the following products.

(i). $103 \times 104$

Ans: On expressing 103 and 104 as the sum of two different numbers, we get,

$103 \times 104 = \left( {100 + 3} \right)\left( {100 + 4} \right)$

On writing the above expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {100 + 3} \right)\left( {100 + 4} \right) = {\left( {100} \right)^2} + \left( {3 + 4} \right)\left( {100} \right) + \left( 3 \right)\left( 4 \right)$

On simplifying the above equation further, we get,

$\left( {100 + 3} \right)\left( {100 + 4} \right) = 10000 + 7\left( {100} \right) + 12$

$\left( {100 + 3} \right)\left( {100 + 4} \right) = 10000 + 700 + 12$

$\left( {100 + 3} \right)\left( {100 + 4} \right) = 10712$

(ii). $5.1 \times 5.2$

Ans: On expressing $5.1$ and $5.2$ as the sum of two different numbers, we get,

$5.1 \times 5.2 = \left( {5 + 0.1} \right)\left( {5 + 0.2} \right)$

On writing the above expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {5 + 0.1} \right)\left( {5 + 0.2} \right) = {\left( 5 \right)^2} + \left( {0.1 + 0.2} \right)\left( 5 \right) + \left( {0.1} \right)\left( {0.2} \right)$

On simplifying the above equation further, we get,

$\left( {5 + 0.1} \right)\left( {5 + 0.2} \right) = 25 + 0.3\left( 5 \right) + 0.02$

$\left( {5 + 0.1} \right)\left( {5 + 0.2} \right) = 25 + 1.5 + 0.02$

$\left( {5 + 0.1} \right)\left( {5 + 0.2} \right) = 26.52$

(iii). $103 \times 98$

Ans: On expressing 103 and 98 as the sum and difference of two different numbers respectively, we get,

$103 \times 98 = \left( {100 + 3} \right)\left( {100 - 2} \right)$

On writing the above expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {100 + 3} \right)\left( {100 - 2} \right) = {\left( {100} \right)^2} + \left[ {3 + \left( { - 2} \right)} \right]\left( {100} \right) + \left( 3 \right)\left( { - 2} \right)$

On simplifying the above equation further, we get,

$\left( {100 + 3} \right)\left( {100 - 2} \right) = 10000 + \left( {3 - 2} \right)\left( {100} \right) - 6$

$\left( {100 + 3} \right)\left( {100 - 2} \right) = 10000 + 100 - 6$

$\left( {100 + 3} \right)\left( {100 - 2} \right) = 10094$

(iv). $9.7 \times 9.8$

Ans: On expressing $9.7$ and $9.8$ as the difference of two different numbers, we get,

$9.7 \times 9.8 = \left( {10 - 0.3} \right)\left( {10 - 0.2} \right)$

On writing the above expression in the form of the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$, we get,

$\left( {10 - 0.3} \right)\left( {10 - 0.2} \right) = {\left( {10} \right)^2} + \left[ {\left( { - 0.3} \right) + \left( { - 0.2} \right)} \right]\left( {10} \right) + \left( { - 0.3} \right)\left( { - 0.2} \right)$

On simplifying the above equation further, we get,

$\left( {10 - 0.3} \right)\left( {10 - 0.2} \right) = 100 + \left[ { - 0.3 - 0.2} \right]\left( {10} \right) + 0.06$

$\left( {10 - 0.3} \right)\left( {10 - 0.2} \right) = 100 + \left( { - 0.5} \right)\left( {10} \right) + 0.06$

$\left( {10 - 0.3} \right)\left( {10 - 0.2} \right) = 100 - 5 + 0.06$

$\left( {10 - 0.3} \right)\left( {10 - 0.2} \right) = 95.06$

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

Opting for the NCERT solutions for Ex 9.5 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.5 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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