# How do you factor 8t^3 -27?

Jun 10, 2016

$\left(2 t - 3\right) \left(4 {t}^{2} + 6 t + 9\right)$

#### Explanation:

This is a $\textcolor{b l u e}{\text{difference of cubes}}$ which is factorised.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here $8 {t}^{3} = {\left(2 t\right)}^{3} \Rightarrow a = 2 t$

and $27 = {\left(3\right)}^{3} \Rightarrow b = 3$

Substitute these into the right side gives.

$8 {t}^{3} - 27 = \left(2 t - 3\right) \left(4 {t}^{2} + 6 t + 9\right)$