# Expand the following

Question:

Expand the following

(i) $(3 a-2 b)^{3}$

(ii) $\left(\frac{1}{x}+\frac{y}{3}\right)^{3}$

(iii) $\left(4-\frac{1}{3 x}\right)^{3}$

Solution:

(i) $(3 a-2 b)^{3}=(3 a)^{3}+(-2 b)^{3}+3(3 a)(-2 b)(3 a-2 b)$

[using identity, $(a-b)^{3}=a^{3}-b^{3}+3 a(-b)(a-b)$ ]

$=27 a^{3}-8 b^{3}-18 a b(3 a-2 b)$

$=27 a^{3}-8 b^{3}-54 a^{2} b+36 a b^{2}$

$=27 a^{3}-54 a^{2} b+36 a b^{2}-8 b^{3}$

(ii) $\left(\frac{1}{x}+\frac{y}{3}\right)^{3}=\left(\frac{1}{x}\right)^{3}+\left(\frac{y}{3}\right)^{3}+3\left(\frac{1}{x}\right)\left(\frac{y}{3}\right)\left(\frac{1}{x}+\frac{y}{3}\right)$

[using identity, $\left.(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right]$

$=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x}\left(\frac{1}{x}+\frac{y}{3}\right)=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3 x}$

(iii) $\left(4-\frac{1}{3 x}\right)^{3}=(4)^{3}+\left(-\frac{1}{3 x}\right)^{3}+3(4)\left(-\frac{1}{3 x}\right)\left(4-\frac{1}{3 x}\right)$

[using identity, $\left.(a-b)^{3}=a^{3}-b^{3}+3 a(-b)(a-b)\right]$

$=64-\frac{1}{27 x^{3}}-\frac{4}{x}\left(4-\frac{1}{3 x}\right)$

$=64-\frac{1}{27 x^{3}}-\frac{16}{x}+\frac{4}{3 x^{2}}$