# How do you factor x^3-27?

Jun 3, 2015

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

Therefore :

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

To convince ourself, we can use Horner's scheme :

Firstly, we need to find a $x$-value for which ${x}^{3} - 27 = 0$.
Here, it's pretty obvious : $x = 3$.

Then, we can use the scheme :

Therefore, ${x}^{3} - 27 = \left(x - 3\right) \left({x}^{2} + 3 x + 9\right)$.

(We can't factorise ${x}^{2} + 3 x + 9$, because $\Delta = {b}^{2} - 4 a c = 9 - 4 \cdot 1 \cdot 9 = - 27 < 0$).