How do you factor a^6+125?

Nov 20, 2015

${a}^{6} + 125 = \left({a}^{2} + 5\right) \left({a}^{4} - 5 {a}^{2} + 25\right)$

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

Explanation:

First use the sum of cubes identity:

${A}^{3} + {B}^{3} = \left(A + B\right) \left({A}^{2} - A B + {B}^{2}\right)$

with $A = {a}^{2}$ and $B = 5$ to find:

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

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

Neither ${a}^{2} + 5$ nor ${a}^{4} - 5 {a}^{2} + 25$ have linear factors with Real coefficients, but ${a}^{4} - 5 {a}^{2} + 25$ will have quadratic factors with Real coefficients.

Consider:

$\left({a}^{2} + k a + 5\right) \left({a}^{2} - k a + 5\right) = {a}^{4} + \left(10 - {k}^{2}\right) {a}^{2} + 25$

If we let $k = \sqrt{15}$ then:

$\left({a}^{2} + k a + 5\right) \left({a}^{2} - k a + 5\right) = {a}^{4} + \left(10 - 15\right) {a}^{2} + 25 = {a}^{4} - 5 {a}^{2} + 25$

So:

${a}^{4} - 5 {a}^{2} + 25 = \left({a}^{2} + \sqrt{15} a + 5\right) \left({a}^{2} - \sqrt{15} a + 5\right)$