# How do you factor 8m³ - 27n³?

Jan 23, 2016

#### Answer:

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

#### Explanation:

Both of these terms are cubed term:

• $8 {m}^{3} = {\left(2 m\right)}^{3}$
• $27 {n}^{3} = {\left(3 n\right)}^{3}$

Because of this, this expression is a difference of cubes.

This is a fairly common pattern that can be factored as

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

Since $8 {m}^{3} - 27 {n}^{3}$ can be expressed as ${\left(2 m\right)}^{3} - {\left(3 n\right)}^{3}$, we can factor this as if $a = 2 m$ and $b = 3 n$.

$8 {m}^{3} - 27 {n}^{3}$

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

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

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