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

Feb 3, 2017

${x}^{6} + 7 {x}^{3} - 8 = \left(x + 2\right) \left({x}^{2} - 2 x + 4\right) \left(x - 1\right) \left({x}^{2} + x + 1\right)$

#### Explanation:

The sum of cubes identity can be written:

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

The difference of cubes identity can be written:

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

So we find:

${x}^{6} + 7 {x}^{3} - 8 = \left({x}^{3} + 8\right) \left({x}^{3} - 1\right)$

$\textcolor{w h i t e}{{x}^{6} + 7 {x}^{3} - 8} = \left({x}^{3} + {2}^{3}\right) \left({x}^{3} - {1}^{3}\right)$

$\textcolor{w h i t e}{{x}^{6} + 7 {x}^{3} - 8} = \left(x + 2\right) \left({x}^{2} - 2 x + 4\right) \left(x - 1\right) \left({x}^{2} + x + 1\right)$