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

Exercise 9.3

1. Carry out the multiplication of the expressions in each of the following pairs.

i. $4p$, $q + r$

Ans: The product of given terms is, 

$\left( {4p} \right) \times \left( {q + r} \right) = 4p \times \left( q \right) + 4p \times \left( r \right)$

$= 4pq + 4pr$

ii. $ab$,$a - b$

Ans: The product of given terms is, 

$\left( {ab} \right) \times \left( {a - b} \right) = \left( {ab} \right) \times a - \left( {ab} \right) \times b$

$= {a^2}b - a{b^2}$

iii. $a + b$,$7{a^2}{b^2}$

Ans: The product of given terms is, 

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

$= 7{a^3}{b^2} + 7{a^2}{b^3}$

iv. ${a^2} - 9$,$4a$

Ans:  The product of given terms is, 

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

$= 4{a^2} + \left( { - 36a} \right)$

$= 4{a^2} - 36a$

v. $pq + qr + rp$,$0$

Ans:  The product of given terms is, 

$\left( {pq + qr + rp} \right) \times \left( 0 \right) = \left( {pq \times 0} \right) + \left( {qr \times 0} \right) + \left( {rp \times 0} \right)$

$= 0 + 0 + 0 = 0$

2. Complete the following table.

Ans: The Complete table is,

3. Find the product.

i. $\left( {{a^2}} \right) \times \left( {2{a^{22}}} \right) \times \left( {4{a^{26}}} \right)$

Ans:  The required product is,

$\left( {{a^2}} \right) \times \left( {2{a^{22}}} \right) \times \left( {4{a^{26}}} \right) = 2 \times 4 \times {a^2} \times {a^{22}} \times {a^{26}}$

$= 8{a^{50}}$

ii. $\left( {\dfrac{2}{3}xy} \right) \times \left( { - \dfrac{9}{{10}}{x^2}{y^2}} \right)$

Ans:  The required product is,

$\left( {\dfrac{2}{3}x} \right) \times \left( { - \dfrac{9}{{10}}{x^2}{y^2}} \right) = \dfrac{2}{3} \times \left( { - \dfrac{9}{{10}}} \right) \times x \times {x^2} \times y \times {y^2}$

$= \left( { - \dfrac{3}{5}} \right) \times x \times {x^2} \times y \times {y^2}$

$= \left( { - \dfrac{3}{5}} \right){x^3}{y^3}$

iii. \[\left( { - \dfrac{{10}}{3}p{q^3}} \right) \times \left( {\dfrac{6}{5}{p^3}q} \right)\]

Ans: The required product is,

$\left( { - \dfrac{{10}}{3}p{q^3}} \right) \times \left( {\dfrac{6}{5}{p^3}q} \right) = - \dfrac{{10}}{3} \times \dfrac{6}{5} \times p \times {p^3} \times {q^3} \times q$

$= - 2 \times 2 \times {p^4} \times {q^4}$

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

iv. $x \times {x^2} \times {x^3} \times {x^4}$

Ans: The required product is,

$ x \times {x^2} \times {x^3} \times {x^4} = {x^{1 + 2 + 3 + 4}}$

$= {x^{10}}$

4. Do as follows.

a. Simplify $3x\left( {4x - 5} \right) + 3$ and find its values for (i)$x = 3$, (ii) $x = \dfrac{1}{2}$.

Ans:  Solve the expression, $3x\left( {4x - 5} \right) + 3$. Multiply the terms in the parenthesis with $3x$.

$3x\left( {4x - 5} \right) + 3 = 3x\left( {4x} \right) - 3x\left( 5 \right) + 3$

$= 12{x^2} - 15x + 3$

i. $x = 3$

Ans:  Substitute $x$ as $3$ in $12{x^2} - 15x + 3$ and simplify.

$12{\left( 3 \right)^2} - 15\left( 3 \right) + 3 = 12\left( 9 \right) - 45 + 3$

$= 108 - 45 + 3$

$= 66$

ii. $x = \dfrac{1}{2}$

Ans:  Substitute $x$ as $\dfrac{1}{2}$ in $12{x^2} - 15x + 3$ and simplify by taking LCM.

$12{\left( {\dfrac{1}{2}} \right)^2} - 15\left( {\dfrac{1}{2}} \right) + 3 = 12 \times \left( {\dfrac{1}{4}} \right) - \dfrac{{15}}{2} + 3$

$= 3 - \dfrac{{15}}{2} + 3$

$ = 6 - \dfrac{{15}}{2}$

$ = \dfrac{{12 - 15}}{2}$

$ =  - \dfrac{3}{2}$

b. Simplify $a\left( {{a^2} + a + 1} \right) + 5$and find its values for the given values of $a$.

Ans: Solve the expression, $a\left( {{a^2} + a + 1} \right) + 5$. Multiply the terms in the parenthesis with $a$ and simplify.

$ a\left( {{a^2} + a + 1} \right) + 5 = a \times \left( {{a^2}} \right) + a \times a + a \times 1 + 5$

$= {a^3} + {a^2} + a + 5$

i. $a = 0$

Ans:  Substitute $a$ as $0$ in ${a^3} + {a^2} + a + 5$ and simplify.

${a^3} + {a^2} + a + 5 = {\left( 0 \right)^3} + {\left( 0 \right)^2} + \left( 0 \right) + 5$ 

$= 5$

ii. $a = 1$

Ans: Substitute $a$ as $1$ in ${a^3} + {a^2} + a + 5$ and simplify.

${a^3} + {a^2} + a + 5 = {\left( 1 \right)^3} + {\left( 1 \right)^2} + \left( 1 \right) + 5$

$= 1 + 1 + 1 + 5$

$= 8$

iii. $a =  - 1$

Ans: Substitute $a$ as $0$ in ${a^3} + {a^2} + a + 5$ and simplify.

${a^3} + {a^2} + a + 5 = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 5$

$=  - 1 + 1 - 1 + 5$

$= 4$

5. Do as follows.

i. Add: $p\left( {p - q} \right)$,$q\left( {q - r} \right)$, and $r\left( {r - p} \right)$.

Ans: Simplify the first expression, $p\left( {p - q} \right) = {p^2} - pq$.

Simplify the second expression, $q\left( {q - r} \right) = {q^2} - qr$.

Simplify the third expression, $r\left( {r - p} \right) = {r^2} - rp$.

Add the three expressions obtained.

$\begin{array}{*{20}{c}}  {}&{{p^2}}& - &{pq}&{}&{}&{}&{}&{}&{}&{}&{} \\   {}&{}&{}&{}& + &{{q^2}}& - &{qr}&{}&{}&{}&{} \\    + &{}&{}&{}&{}&{}&{}&{}& + &{{r^2}}& - &{pq} \\ \hline  {}&{{p^2}}& - &{pq}& + &{{q^2}}& - &{qr}& + &{{r^2}}& - &{pq} \end{array}$

Hence, the the sum of the given expression is ${p^2} - pq + {q^2} - qr + {r^2} - pq$.

ii. Add: $2x\left( {z - x - y} \right)$ and $2y\left( {z - y - x} \right)$.

Ans: Simplify the first expression,$2x\left( {z - x - y} \right)$.

$2x\left( {z - x - y} \right) = 2x \times z - 2x \times x - 2x \times y$

$= 2xz - 2{x^2} - 2xy$

Simplify the second expression, $2y\left( {z - y - x} \right)$.

$2y\left( {z - y - x} \right) = 2y \times z - 2y \times y - 2y \times x$

$= 2yz - 2{y^2} - 2yx$

Now, add the two expressions obtained.

$\,\,2xz - 2{x^2} - 2xy \\  \underline {\left(  +  \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2xy - 2{y^2} - 2yz} \\  2xz - 2{x^2} - 4xy + 2yz - 2{y^2}$

Hence, the sum of the given expression is $2xz - 2{x^2} - 4xy + 2yz - 2{y^2}$.

iii. Subtract $3l\left( {l - 4m + 5n} \right)$ from $4l\left( {10n - 3m + 2l} \right)$.

Ans: Simplify the first expression, $3l\left( {l - 4m + 5n} \right)$.

$3l\left( {l - 4m + 5n} \right) = 3l \times l - 3l \times 4m + 3l \times 5n$

$= 3{l^2} - 12lm + 15\ln$

Simplify the second expression, $4l\left( {10n - 3m + 2l} \right)$.

$ 4l\left( {10n - 3m + 2l} \right) = 4l \times 10n - 4l \times 3m + 4l \times 2l$

$= 40\ln  - 12lm + 8{l^2}$

Subtract the obtained expression.

$\,\,\,\,\,\,\,\,40\ln  - 12lm + 8{l^2} \\  \underline {\left(  -  \right)15\ln  - 12lm + 3{l^2}\,\,\,\,}  \\  \,\,\,\,\,\,\,\,25\ln  + 0 + \,\,\,\,\,\,5{l^2}$

The difference is $5{l^2} + 25\ln $.

iv. Subtract: $3a\left( {a + b + c} \right) - 2b\left( {a - b + c} \right)$ from $4c\left( { - a + b + c} \right)$.

Ans:  Simplify the first expression, $3a\left( {a + b + c} \right) - 2b\left( {a - b + c} \right)$.

$3a\left( {a + b + c} \right) - 2b\left( {a - b + c} \right) = 3a \times \left( a \right) + 3a \times \left( b \right) + 3a \times \left( c \right) - 2b\left( a \right) + 2b\left( b \right) - 2b\left( c \right)$

$= 3{a^2} + 3ab + 3ac - 2ab + 2{b^2} - 2bc$

$= 3{a^2} + ab + 3ac + 2{b^2} - 2bc$

Simplify the second expression, $4c\left( { - a + b + c} \right)$.

$4c\left( { - a + b + c} \right) = 4c \times \left( { - a} \right) + 4c \times \left( b \right) + 4c \times c$

$=  - 4ac + 4cb + 4{c^2}$

Subtract the obtained expression.

$\,\,\,\,\,\,\,\,-4ac + 4cb + 4{c^2} \\  \underline {\left(  -  \right)3ac - 2bc + 3{a^2} + 2{b^2} + ab \,\,\,\,}  \\  \,\,\,- 7ac + 6bc + 4{c^2} - 3{a^2} - 2{b^2} - ab$

The simplified expression is, $ - 7ac + 6bc + 4{c^2} - 3{a^2} - 2{b^2} - ab$.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities (Ex 9.3) Exercise 9.3

Opting for the NCERT solutions for Ex 9.3 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 9.3 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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