# How do you factor x^6y^3 + y^9?

Feb 25, 2017

${x}^{6} {y}^{3} + {y}^{9} = {y}^{3} \left({x}^{2} + {y}^{2}\right) \left({x}^{2} - \sqrt{3} x y + {y}^{2}\right) \left({x}^{2} + \sqrt{3} x y + {y}^{2}\right)$

#### Explanation:

The sum of cubes identity can be written:

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

So we find:

${x}^{6} {y}^{3} + {y}^{9} = {y}^{3} \left({x}^{6} + {y}^{6}\right)$

$\textcolor{w h i t e}{{x}^{6} {y}^{3} + {y}^{9}} = {y}^{3} \left({\left({x}^{2}\right)}^{3} + {\left({y}^{2}\right)}^{3}\right)$

$\textcolor{w h i t e}{{x}^{6} {y}^{3} + {y}^{9}} = {y}^{3} \left({x}^{2} + {y}^{2}\right) \left({x}^{4} - {x}^{2} {y}^{2} + {y}^{4}\right)$

Then note that:

$\left({x}^{2} - a x y + {y}^{2}\right) \left({x}^{2} + a x y + {y}^{2}\right) = {x}^{4} + \left(2 - {a}^{2}\right) {x}^{2} {y}^{2} + {y}^{4}$

So putting $a = \sqrt{3}$ we find:

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

Putting it all together:

${x}^{6} {y}^{3} + {y}^{9} = {y}^{3} \left({x}^{2} + {y}^{2}\right) \left({x}^{2} - \sqrt{3} x y + {y}^{2}\right) \left({x}^{2} + \sqrt{3} x y + {y}^{2}\right)$