Question
Factorize the following expressions:
$\frac{\text{x}^3}{216}=8\text{y}^3$
$\frac{\text{x}^3}{216}=8\text{y}^3$
✓
Answer
$\frac{\text{x}^3}{216}-8\text{y}^3$
$=\frac{\text{x}^3}{6}-(2\text{y})^3$
$=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\Big(\frac{\text{x}}{6}\Big)^2+\frac{\text{x}}{6}\times2\text{y}+(2\text{y})^2\Big)$
$\therefore\big[\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2)\big]$
$=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}^2}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
$\therefore\frac{\text{x} ^3}{216}-8\text{y} ^3=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
$=\frac{\text{x}^3}{6}-(2\text{y})^3$
$=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\Big(\frac{\text{x}}{6}\Big)^2+\frac{\text{x}}{6}\times2\text{y}+(2\text{y})^2\Big)$
$\therefore\big[\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2)\big]$
$=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}^2}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
$\therefore\frac{\text{x} ^3}{216}-8\text{y} ^3=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
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