# How do you factor 27x^3 - 8y^3 ?

Jan 15, 2016

$27 {x}^{3} - 8 {y}^{3} = {\left(3 x\right)}^{3} - {\left(2 y\right)}^{3} = \textcolor{red}{\left(3 x - 2 y\right)} \textcolor{red}{\left(9 {x}^{2} + 6 x y + 4 {y}^{2}\right)}$

#### Explanation:

$27 {x}^{3} - 8 {y}^{3}$ fits the form of the difference of cubes, ${a}^{3} - {b}^{3}$, where $a = 3 x$ and $b = 2 y$.

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

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

Substitute the known values for $a$ and $b$.

${\left(3 x\right)}^{3} - {\left(2 y\right)}^{3} = \left(3 x - 2 y\right) \left({\left(3 x\right)}^{2} + \left(3 x \cdot 2 y\right) + {\left(2 y\right)}^{2}\right)$

Simplify.

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