Question
Factorise : $\frac{(8 a)^3}{27}-\frac{b^3}{8}$
✓
Answer
$ \frac{(8 a)^3}{27}-\frac{b^3}{8}$
$ =\left(\frac{2 a}{3}\right)^3-\left(\frac{b}{2}\right)^3$
$ =\left(\frac{2 a}{3}-\frac{b}{2}\right)\left[\left(\frac{2 a}{3}\right)^2+\frac{2 a}{3} \times \frac{b}{2}+\left(\frac{b}{2}\right)^2\right]$
$ {\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right] }$
$ =\left(\frac{2 a}{3}-\frac{b}{2}\right)\left[\frac{4 a^2}{9}+\frac{a b}{3}+\frac{b^2}{4}\right]$
$ =\left(\frac{2 a}{3}\right)^3-\left(\frac{b}{2}\right)^3$
$ =\left(\frac{2 a}{3}-\frac{b}{2}\right)\left[\left(\frac{2 a}{3}\right)^2+\frac{2 a}{3} \times \frac{b}{2}+\left(\frac{b}{2}\right)^2\right]$
$ {\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right] }$
$ =\left(\frac{2 a}{3}-\frac{b}{2}\right)\left[\frac{4 a^2}{9}+\frac{a b}{3}+\frac{b^2}{4}\right]$
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