# How do you factor x^3-729 ?

Dec 10, 2015

$\left(x - 9\right) \left({x}^{2} + 9 x + 81\right)$

#### Explanation:

You have to observe that $729 = {9}^{3}$, and so ${x}^{3} - 729$ is a difference of cubes.

In general, you have that

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

and $\left({a}^{2} + a b + {b}^{2}\right)$ is no longer factorizable (it is also known as "false square", because it resembles ${\left(a + b\right)}^{2}$, but the mixed product is halved).

In your case, $a = x$ and $b = 9$, so we have

${x}^{3} - {9}^{3} = \left(x - 9\right) \left({x}^{2} + 9 x + 81\right)$