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Question:

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(i) $(3 x+2)^{3}$

(ii) $\left(3 a+\frac{1}{4 b}\right)^{3}$

(iii) $\left(1+\frac{2}{3} a\right)^{3}$

 

Solution:

(i) $(3 x+2)^{3}=(3 x)^{3}+3 \times(3 x)^{2} x 2+3 \times 3 x \times(2)^{2}+(2)^{3}$

$=27 x^{3}+54 x^{2}+36 x+8$

(ii) $\left(3 a+\frac{1}{4 b}\right)^{3}=(3 a)^{3}+\left(\frac{1}{4 b}\right)^{3}+3(3 a)^{2}\left(\frac{1}{4 b}\right)+3(3 a)\left(\frac{1}{4 b}\right)^{2}$

$=27 a^{3}+\frac{1}{64 b^{3}}+\frac{27 a^{2}}{4 b}+\frac{9 a}{16 b^{2}}$

(iii) $\left(1+\frac{2}{3} a\right)^{3}=\left(\frac{2}{3} a\right)^{3}+3 \times\left(\frac{2}{3} a\right)^{2} \times 1+3 a \frac{2}{3} a \times(1)^{2}+(1)^{3}$

$=\frac{8}{27} a^{3}+\frac{4}{3} a^{2}+2 a+1$

 

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