# NCERT Solutions Class 8 Mathematics Solutions for Exercise 9.4 in Chapter 9 - Algebraic Expressions and Identities

Question 6 Exercise 9.4

Simplify.

(i) \left(x^{2}-5\right)(x+5)+25

(ii) \left(a^{2}+5\right)\left(b^{3}+3\right)+5

(iii) \left(t+s^{2}\right)\left(t^{2}-s\right)

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

(v) (x + y)(2x + y) + (x + 2y)(x – y)

(vi) (x+y)\left(x^{2}-x y+y^{2}\right)

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

(viii) (a + b + c)(a + b – c)

(i)

\begin{array}{l} \left(x^{2}-5\right)(x+5)+25 \\ =x^{3}+5 x^{2}-5 x-25+25 \\ =x^{3}+5 x^{2}-5 x \end{array}

(ii)

\begin{array}{l} \left(a^{2}+5\right)\left(b^{3}+3\right)+5 \\ =a^{2} b^{3}+3 a^{2}+5 b^{3}+15+5 \\ =a^{2} b^{3}+5 b^{3}+3 a^{2}+20 \end{array}

(iii)

\begin{aligned} &\left(t+s^{2}\right)\left(t^{2}-s\right)\\ &=t\left(t^{2}-s\right)+s^{2}\left(t^{2}-s\right)\\ &=t^{3}-s t+s^{2} t^{2}-s^{3}\\ &=t^{3}-s^{3}-s t+s^{2} t^{2} \end{aligned}

iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

= (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

=(ac - ad + bc - bd) + (ac + ad - bc - bd) + (2ac + 2bd)

= ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd

= 4ac

v) (x + y)(2x + y) + (x + 2y)(x – y)

\begin{array}{l} =2 x^{2}+x y+2 x y+y^{2}+x^{2}-x y+2 x y-2 y^{2} \\ =3 x^{2}+4 x y-y^{2} \end{array}

(vi)

\begin{array}{l} (\mathrm{x}+\mathrm{y})\left(\mathrm{x}^{2}-\mathrm{xy}+\mathrm{y}^{2}\right) \\ =\mathrm{x}^{3}-\mathrm{x}^{2} \mathrm{y}+\mathrm{xy}^{2}+\mathrm{x}^{2} \mathrm{y}-\mathrm{xy}^{2}+\mathrm{y}^{3} \\ =\mathrm{x}^{3}+\mathrm{y}^{3} \end{array}

vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

\begin{array}{l} =2.25 \mathrm{x}^{2}+6 \mathrm{xy}+4.5 \mathrm{x}-6 \mathrm{xy}-16 \mathrm{y}^{2}-12 \mathrm{y}-4.5 \mathrm{x}+12 \mathrm{y} \\ =2.25 \mathrm{x}^{2}-16 \mathrm{y}^{2} \end{array}

viii) (a + b + c)(a + b – c)

\begin{array}{l} =a^{2}+a b-a c+a b+b^{2}-b c+a c+b c-c^{2} \\ =a^{2}+b^{2}-c^{2}+2 a b \end{array}

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