Question
Factorise : $27 p^3-\frac{1}{216}-\frac{9}{2} p^2+\frac{1}{4} p$.
✓
Answer
$27 p^3-\frac{1}{216}-\frac{9}{2} p^2+\frac{1}{4} p$
$=(3 p)^3-\left(\frac{1}{6}\right)^3-3(3 p)\left(\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
$=\left(3 p-\frac{1}{6}\right)^3$
(Using Identity $(a-b)^3=a^3-b^3-3 a b(a-b)$ )
$=\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
$=(3 p)^3-\left(\frac{1}{6}\right)^3-3(3 p)\left(\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
$=\left(3 p-\frac{1}{6}\right)^3$
(Using Identity $(a-b)^3=a^3-b^3-3 a b(a-b)$ )
$=\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
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