Question

Factorize: $27p^3-\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4}p$

Answer 1

Given:

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

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

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

We know that,

$a^3-3a^{2}b+3a{b}^2-b^3=(a-b{)}^3$

So lets compare,

$({\left(\frac{1}{6}\right)}^3-3{\left(\frac{1}{6}\right)}^2(3p)+3\left(\frac{1}{6}\right)(3p{)}^2-\left(3p{)}^3\right)$ and $a^3-3a^{2}b+3a{b}^2-b^3$

$a^3={\left(\frac{1}{6}\right)}^3$

$\Rightarrow a=\frac{1}{6}$ ---(2) and

$b^3=(3p{)}^3$

$\Rightarrow b=3p$ ---(3)

Now,

$a^3-3a^{2}b+3a{b}^2-b^3=(a-b{)}^3$

$\Rightarrow ({\left(\frac{1}{6}\right)}^3-3{\left(\frac{1}{6}\right)}^2(3p)+3\left(\frac{1}{6}\right)(3p{)}^2-\left(3p{)}^3\right)={(\frac{1}{6}-3p)}^3$

$\Rightarrow -({\left(\frac{1}{6}\right)}^3-3{\left(\frac{1}{6}\right)}^2(3p)+3\left(\frac{1}{6}\right)(3p{)}^2-\left(3p{)}^3\right)=-{(\frac{1}{6}-3p)}^3$

$\Rightarrow -({\left(\frac{1}{6}\right)}^3-3{\left(\frac{1}{6}\right)}^2(3p)+3\left(\frac{1}{6}\right)(3p{)}^2-\left(3p{)}^3\right)={(3p-\frac{1}{6})}^3$

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