8.(v) Factorise the following:

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

Answers (1)
R Riya

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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