Question
Factorise:
$8\text{x}^3-\frac{1}{27\text{y}^3}$

Answer

We know that:
Since a3 - b3 = (a - b)(a2 + a × b + b2)
Let us rewrite
$8\text{x}^3-\frac{1}{27\text{y}^3}$
$=(2\text{x})^3-\Big(\frac{1}{3\text{y}}\Big)^3$
$=\Big(2\text{x}-\frac{1}{3\text{y}}\Big)\bigg[(2\text{x})^2+2\text{x}\times\frac{1}{3\text{y}}+\Big(\frac{1}{3\text{y}}\Big)^2\bigg]$
$=\Big(2\text{x}-\frac{1}{3\text{y}}\Big)\Big(4\text{x}^2+\frac{2\text{x}}{3\text{y}}+\frac{1}{9\text{y}^2}\Big)$

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