# How do you factor (x+y)^3 - 9(x^9 - y^9) ?

Aug 9, 2016

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

#### Explanation:

First, we will simplify $\left({x}^{9} - {y}^{9}\right)$, since this is a difference of cubes. Differences of cubes fit the identity: ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$.

Thus:

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

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

Now $\left({x}^{3} - {y}^{3}\right)$ can be factored in the same way:

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

This is really the furthest this can be factored.