Question
Factorise:$216\text{x}^3+\frac{1}{125}$

Answer

$216\text{x}^3+\frac{1}{125}$We know that:
Since $a^2 + b^3 = (a + b)(a^2 - a \times b + b^2)$
Let us rewrite
$216\text{x}^3+\frac{1}{125}$
$=(6\text{x})^3+\Big(\frac{1}{5}\Big)^3$
$=\Big(6\text{x}+\frac{1}{5}\Big)\bigg[(6\text{x})^2-6\text{x}\times\frac{1}{5}+\Big(\frac{1}{5}\Big)^2\bigg]$
$=\Big(6\text{x}+\frac{1}{5}\Big)\Big(36\text{x}^2-\frac{6\text{x}}{5}+\frac{1}{25}\Big)$

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