How do you factor 125x^9 + 729y^6?

Jan 2, 2016

Use the sum of cubes identity to find:

$125 {x}^{9} + 729 {y}^{6} = \left(5 {x}^{3} + 9 {y}^{2}\right) \left(25 {x}^{6} - 45 {x}^{3} {y}^{2} + 81 {y}^{4}\right)$

Explanation:

Both $125 {x}^{9} = {\left(5 {x}^{3}\right)}^{3}$ and $729 {y}^{6} = {\left(9 {y}^{2}\right)}^{3}$ are perfect cubes.

The sum of cubes identity can be written:

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

We can use this with $a = 5 {x}^{3}$ and $b = 9 {y}^{2}$ to find:

$125 {x}^{9} + 729 {y}^{6}$

$= {\left(5 x\right)}^{3} + {\left(9 {y}^{2}\right)}^{3}$

$= \left(5 {x}^{3} + 9 {y}^{2}\right) \left({\left(5 {x}^{3}\right)}^{2} - \left(5 {x}^{3}\right) \left(9 {y}^{2}\right) + {\left(9 {y}^{2}\right)}^{2}\right)$

$= \left(5 {x}^{3} + 9 {y}^{2}\right) \left(25 {x}^{6} - 45 {x}^{3} {y}^{2} + 81 {y}^{4}\right)$

That's as far as we can go with Real coefficients.

If you allow Complex coefficients then it can be further factored as:

$= \left(5 {x}^{3} + 9 {y}^{2}\right) \left(5 {x}^{3} + 9 \omega {y}^{2}\right) \left(5 {x}^{3} + 9 {\omega}^{2} {y}^{2}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.