Question

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

Answer 1

The expression $27p^3-\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4}p$ can be written as ${\left(3p\right)}^3-{\left(\frac{1}{6}\right)}^3-3{\left(3p\right)}^2\left(\frac{1}{6}\right)+3\left(3p\right){\left(\frac{1}{6}\right)}^2$ ….[1]

We know the Algebraic identity, ${\left(a-b\right)}^3=a^3-b^3-3ab\left(a-b\right)$

$=a^3-b^3-3a^{2}b+3a{b}^2$ …[2]

By comparing the expressions [1] and [2], we can say that,

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

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

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