Question

Factorize $8p^3-\frac{12}{5}{p}^2+\frac{6}{25}p-\frac{1}{125}$.

Answer 1

Factorize$\mathbf{8}{\mathit{p}}^{\mathbf{3}}\mathbf{-}\frac{\mathbf{12}}{\mathbf{5}}{\mathit{p}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{6}}{\mathbf{25}}\mathit{p}\mathbf{-}\frac{\mathbf{1}}{\mathbf{125}}$.

We can write the given expression as

$8p^3-\frac{12}{5}{p}^2+\frac{6}{25}p-\frac{1}{125}={\left(2p\right)}^3-3{\left(2p\right)}^2\times \frac{1}{5}+3\left(2p\right)\times {\left(\frac{1}{5}\right)}^2-{\left(\frac{1}{5}\right)}^3$

$={\left(2p-\frac{1}{5}\right)}^3$ $\left[\therefore {\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2-b^3\right]$

$=\left(2p-\frac{1}{5}\right)\left(2p-\frac{1}{5}\right)\left(2p-\frac{1}{5}\right)$

Hence, the factored expression is $\left(2p-\frac{1}{5}\right)\left(2p-\frac{1}{5}\right)\left(2p-\frac{1}{5}\right)$.