# How do you factor the expression x^9-27y^6?

Apr 7, 2018

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

#### Explanation:

Given:

${x}^{9} - 27 {y}^{6}$

Note that both ${x}^{9} = {\left({x}^{3}\right)}^{3}$ and $27 {y}^{6} = {\left(3 {y}^{2}\right)}^{3}$ are perfect cubes.

So we can factor the given expression as a difference of cubes:

${A}^{3} - {B}^{3} = \left(A - B\right) \left({A}^{2} + A B + {B}^{2}\right)$

with $A = {x}^{3}$ and $B = 3 {y}^{2}$ as follows:

${x}^{9} - 27 {y}^{6} = {\left({x}^{3}\right)}^{3} - {\left(3 {y}^{2}\right)}^{3}$

$\textcolor{w h i t e}{{x}^{9} - 27 {y}^{6}} = \left({x}^{3} - 3 {y}^{2}\right) \left({\left({x}^{3}\right)}^{2} + \left({x}^{3}\right) \left(3 {y}^{2}\right) + {\left(3 {y}^{2}\right)}^{2}\right)$

$\textcolor{w h i t e}{{x}^{9} - 27 {y}^{6}} = \left({x}^{3} - 3 {y}^{2}\right) \left({x}^{6} + 3 {x}^{3} {y}^{2} + 9 {y}^{4}\right)$