How do you factor x^6-124x^3-125?

Feb 17, 2017

${x}^{6} - 124 {x}^{3} - 125 = \left(x + 1\right) \left({x}^{2} - x + 1\right) \left(x - 5\right) \left({x}^{2} + 5 x + 25\right)$

Explanation:

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

Hence we find:

${x}^{6} - 124 {x}^{3} - 125 = \left({x}^{6} - 125 {x}^{3}\right) + \left({x}^{3} - 125\right)$

$\textcolor{w h i t e}{{x}^{6} - 124 {x}^{3} - 125} = {x}^{3} \left({x}^{3} - 125\right) + 1 \left({x}^{3} - 125\right)$

$\textcolor{w h i t e}{{x}^{6} - 124 {x}^{3} - 125} = \left({x}^{3} + 1\right) \left({x}^{3} - 125\right)$

$\textcolor{w h i t e}{{x}^{6} - 124 {x}^{3} - 125} = \left({x}^{3} + {1}^{3}\right) \left({x}^{3} - {5}^{3}\right)$

$\textcolor{w h i t e}{{x}^{6} - 124 {x}^{3} - 125} = \left(x + 1\right) \left({x}^{2} - x + 1\right) \left(x - 5\right) \left({x}^{2} + 5 x + 25\right)$