# How do you factor x^6 - 64?

Apr 15, 2015

(x+2)(x-2)(${x}^{4}$ +4${x}^{2}$+16)

Rewrte the given terms as perfect cubes as follows and use the algebraic identity for the factorisation of difference of two cubes.

${x}^{6}$ -64 = ${\left({x}^{2}\right)}^{3}$ - ${\left(4\right)}^{3}$

=(${x}^{2}$ -4) (${x}^{4}$ +4${x}^{2}$+16)

=(x+2)(x-2)(${x}^{4}$ +4${x}^{2}$+16)

Apr 15, 2015

This is a bit of a tricky question.

If you write ${x}^{6} - 64 = {\left({x}^{2}\right)}^{3} - {4}^{3}$ you can factor it into:

$\left(x + 2\right) \left(x - 2\right) \left({x}^{4} + 4 {x}^{2} + 16\right)$

If you write:
${x}^{6} - 64 = {\left({x}^{3}\right)}^{2} - {8}^{2}$ , then you factor as

$\left({x}^{3} + 8\right) \left({x}^{3} - 8\right)$ which can be further factored as:

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

The second answer is factored into irreducibles (over $\mathbb{R}$).