# NCERT Solutions Class 8 Mathematics Solutions for Exercise 9.5 in Chapter 9 - Algebraic Expressions and Identities

Question 10 Exercise 9.5

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

\text { (ii) }(9 p-5 q)^{2}+180 p q=(9 p+5 q)^{2}

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

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

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

\begin{aligned} &\text { i) } \mathrm{LHS}=(3 \mathrm{x}+7)^{2}-84 \mathrm{x}\\ &\begin{array}{l} =9 x^{2}+42 x+49-84 x \\ =9 x^{2}-42 x+49 \end{array} \end{aligned}

= RHS

LHS = RHS

\begin{aligned} &\text { ii) } \mathrm{LHS}=(9 \mathrm{p}-5 \mathrm{q})^{2}+180 \mathrm{pq}\\ &\begin{aligned} &=81 \mathrm{p}^{2}-90 \mathrm{pq}+25 \mathrm{q}^{2}+180 \mathrm{pq} \\ &=81 \mathrm{p}^{2}+90 \mathrm{pq}+25 \mathrm{q}^{2} \\ \mathrm{RHS} &=(9 \mathrm{p}+5 \mathrm{q})^{2} \\ &=81 \mathrm{p}^{2}+90 \mathrm{pq}+25 \mathrm{q}^{2} \end{aligned} \end{aligned}

LHS = RHS

\begin{aligned} &\text { (iii) } L H S=\left(\frac{4}{3} m-\frac{3}{4} n\right)^{2}+2 m n\\ &=\frac{16}{9} m^{2}+\frac{9}{16} n^{2}-2 m n+2 m n\\ &=\frac{16}{9} m^{2}+\frac{9}{16} n^{2}\\ &=\mathrm{RHS}\\ &\mathrm{LHS}=\mathrm{RHS} \end{aligned}

\begin{aligned} &\text { iv) } \mathrm{LHS}=(4 \mathrm{pq}+3 \mathrm{q})^{2}-(4 \mathrm{pq}-3 \mathrm{q})^{2}\\ &\begin{aligned} &=16 \mathrm{p}^{2} \mathrm{q}^{2}+24 \mathrm{pq}^{2}+9 \mathrm{q}^{2}-16 \mathrm{p}^{2} \mathrm{q}^{2}+24 \mathrm{pq}^{2}-9 \mathrm{q}^{2} \\ &=48 \mathrm{pq}^{2} \\ \mathrm{RHS} &=48 \mathrm{pq}^{2} \\ \mathrm{LHS} &=\mathrm{RHS} \end{aligned} \end{aligned}

\begin{aligned} &\text { v) } \mathrm{LHS}=(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})+(\mathrm{b}-\mathrm{c})(\mathrm{b}+\mathrm{c})+(\mathrm{c}-\mathrm{a})(\mathrm{c}+\mathrm{a})\\ &\begin{array}{l} =a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2} \\ =0 \\ =R H S \end{array} \end{aligned}

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