# How do you factor x^6-7x^3-8?

${x}^{6} - 7 {x}^{3} - 8 \equiv {\left({x}^{3}\right)}^{2} - 7 \left({x}^{3}\right) - 8 \equiv \left({x}^{3} - 8\right) \left({x}^{3} + 1\right) \equiv \left({x}^{3} - {\left(2\right)}^{3}\right) \left({x}^{3} + {\left(1\right)}^{3}\right)$
Recall that, ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$ and ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
$\implies \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) \left(x + 1\right) \left({x}^{2} - x + 1\right)$