# How do you factor x^4-8x?

${x}^{4} - 8 x = x \left({x}^{3} - 8\right) = x \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$.
The last passage is because $\left({x}^{3} - 8\right)$ is a difference of two cubes, that can be factored with this rule:
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$.