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# Factorise the following: (v) 27 p ^3 - 1 / 216 - 9 / 2 p ^2 + 1 / 4 p

8.(v) Factorise the following:

(v)    $27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$

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We can rewrite   $27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$  as

$\Rightarrow 27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$ $= (3p)^3-\left(\frac{1}{6} \right )^3-3.(3p)^2.\frac{1}{6}+3.3p.\left(\frac{1}{6} \right )^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here,  $x=3p \ \ and \ \ y= \frac{1}{6}$

Therefore,

$27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p = \left ( 3p-\frac{1}{6} \right )^3$

$= \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right )$

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