# How do you factor 64v^3-125?

Aug 18, 2016

$\left(4 v - 5\right) \left(16 {v}^{2} + 20 v + 25\right)$

#### Explanation:

This is a $\textcolor{b l u e}{\text{difference of cubes}}$ and , in general, factorises as.

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

Now $64 {v}^{3} = {\left(4 v\right)}^{3} \text{ and } 125 = {\left(5\right)}^{3}$

$\Rightarrow a = 4 v \text{ and } b = 5$

substitute into (A)

$\Rightarrow 64 {v}^{3} - 125 = \left(4 v - 5\right) \left(16 {v}^{2} + 20 v + 25\right)$