# How do you factor x^6+125?

Feb 4, 2017

${x}^{6} + 125 = \left({x}^{2} + 5\right) \left({x}^{2} - \sqrt{15} x + 5\right) \left({x}^{2} + \sqrt{15} x + 5\right)$

#### Explanation:

The sum of cubes identity can be written:

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

Hence we find:

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

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

$\textcolor{w h i t e}{{x}^{6} + 125} = \left({x}^{2} + 5\right) \left({x}^{4} - 5 {x}^{2} + 25\right)$

To factor the remaining quartic, note that:

$\left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right) = {a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}$

So with $a = x$ and $b = \sqrt{5}$, we find:

$\left({x}^{2} - k \sqrt{5} x + 5\right) \left({x}^{2} + k \sqrt{5} x + 5\right) = {x}^{4} + \left(2 - {k}^{2}\right) 5 {x}^{2} + 25$

Equating coefficients, we want:

$\left(2 - {k}^{2}\right) 5 = - 5$

Hence:

$k = \pm \sqrt{3}$

So:

${x}^{4} - 5 {x}^{2} + 25 = \left({x}^{2} - \sqrt{3} \sqrt{5} x + 5\right) \left({x}^{2} + \sqrt{3} \sqrt{5} x + 5\right)$

$\textcolor{w h i t e}{{x}^{4} - 5 {x}^{2} + 25} = \left({x}^{2} - \sqrt{15} x + 5\right) \left({x}^{2} + \sqrt{15} x + 5\right)$

Putting it all together:

${x}^{6} + 125 = \left({x}^{2} + 5\right) \left({x}^{2} - \sqrt{15} x + 5\right) \left({x}^{2} + \sqrt{15} x + 5\right)$

We can quickly determine that none of these quadratics has linear factors with real coefficients since ${x}^{6} + 125$ has no real zeros.