# How do you factor 8m^3 - 125n^3?

##### 2 Answers
May 16, 2018

$\left(2 m - 5 n\right) \left(4 {m}^{2} + 10 m n + 25 {n}^{2}\right)$

#### Explanation:

$\text{this is a "color(blue)"difference of cubes}$

$\text{and factors in general as}$

•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)

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

$125 {n}^{3} = {\left(5 n\right)}^{3} \Rightarrow b = 5 n$

$8 {m}^{3} - 125 {n}^{3} = \left(2 m - 5 n\right) \left({\left(2 m\right)}^{2} + \left(2 m \times 5 n\right) + {\left(5 n\right)}^{2}\right)$

$\textcolor{w h i t e}{\times \times \times \times x} = \left(2 m - 5 n\right) \left(4 {m}^{2} + 10 m n + 25 {n}^{2}\right)$

May 16, 2018

$\left(2 m - 5 n\right) \left(4 {m}^{2} + 10 m n + 25 {n}^{2}\right)$

#### Explanation:

Remember that ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

$8 {m}^{3} - 125 {n}^{3} = {2}^{3} {m}^{3} - {5}^{3} {n}^{3} = {\left(2 m\right)}^{3} - {\left(5 n\right)}^{3}$

Therefore, $a = 2 m , b = 5 n$

Sub in: $\left(2 m - 5 n\right) \left(4 {m}^{2} + 10 m n + 25 {n}^{2}\right)$