# How do you factor a^3 - (a-4)^3?

Feb 15, 2016

$4 \left(3 {a}^{2} - 12 a + 16\right)$

#### Explanation:

This is a difference of cubes, which factors into:

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

Here, we have $x = a$ and $y = \left(a - 4\right)$, which gives us a factorization of

${a}^{3} - {\left(a - 4\right)}^{3}$

$= \left(a - \left(a - 4\right)\right) \left({a}^{2} + a \left(a - 4\right) + {\left(a - 4\right)}^{2}\right)$

We can continue to simplify.

$= \left(a - a + 4\right) \left({a}^{2} + {a}^{2} - 4 a + {a}^{2} - 8 a + 16\right)$

$= 4 \left(3 {a}^{2} - 12 a + 16\right)$

This cannot be factored further (without the help of complex numbers).