# How do you factor (x-3)^3 + (3x-2)^3?

Mar 19, 2018

$= 7 \left(4 x - 5\right) \left({x}^{2} - x + 1\right)$

#### Explanation:

Both terms are cubes and the expression can be factored as:

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

To make the process easier to follow,

let $\left(x - 3\right) = m \mathmr{and} \left(3 x - 2\right) = n$

${m}^{3} + {n}^{3}$

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

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

Simplify:

$= \left(4 x - 5\right) \left({x}^{2} - 6 x + 9 - \left(3 {x}^{2} - 11 x + 6\right) + 9 {x}^{2} - 12 x + 4\right)$

$= \left(4 x - 5\right) \left({x}^{2} - 6 x + 9 - 3 {x}^{2} + 11 x - 6 + 9 {x}^{2} - 12 x + 4\right)$

$= \left(4 x - 5\right) \left(7 {x}^{2} - 7 x + 7\right)$

$= 7 \left(4 x - 5\right) \left({x}^{2} - x + 1\right)$