# How do you factor 125x^6 - 8y^6?

May 11, 2015

It's $\left(\sqrt{5} x + \sqrt{2} y\right) \left(\sqrt{5} x - \sqrt{2} y\right) \left(25 {x}^{4} + 10 {x}^{2} {y}^{2} + 4 {y}^{4}\right)$

Let's write $P = P \left(x , y\right) = 125 {x}^{6} - 8 {y}^{6}$

It's a difference of cubes ($125 {x}^{6} = {\left(5 {x}^{2}\right)}^{3} , 8 {y}^{6} = {\left(2 {y}^{2}\right)}^{3}$)

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

So $P = {P}_{1} \cdot {P}_{2} = \left(5 {x}^{2} - 2 {y}^{2}\right) \cdot \left(25 {x}^{4} + 10 {x}^{2} {y}^{2} + 4 {y}^{4}\right)$

${P}_{1}$ is a difference of squares:

So ${P}_{1} = \left(\sqrt{5} x + \sqrt{2} y\right) \left(\sqrt{5} x - \sqrt{2} y\right)$

${P}_{2}$ has no real roots and it's not product of two polynomials (just try it, it's boring but I can't find a faster way to prove it) so you're done!

May 11, 2015

$125 {x}^{6} - 8 {y}^{6} = {\left(5 {x}^{2}\right)}^{3} - {\left(2 {y}^{2}\right)}^{3}$

Notice that this is of the form $\left({a}^{3} - {b}^{3}\right)$ which can be factored as $\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

So we can write

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

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

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

This factorisation is complete.

For fun let us see what lies beyond integer and rational coefficients ...

The first of these factors, $\left(5 {x}^{2} - 2 {y}^{2}\right)$ does have linear factors, but only ones with irrational coefficients, for example:

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

To check that the second factor, $\left(25 {x}^{4} + 10 {x}^{2} {y}^{2} + 4 {y}^{4}\right)$ has no simpler factors with real coefficients, let us see what factors it does have (this starts to get a little messy).

Let $\omega = - \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right) i$ where $i = \sqrt{- 1}$.

Then ${\omega}^{2} = - \frac{1}{2} - \left(\frac{\sqrt{3}}{2}\right)$ and ${\omega}^{3} = 1$.

Notice also that $\omega + {\omega}^{2} = - 1$.

Then we have:

$\left(5 {x}^{2} - 2 \omega {y}^{2}\right) \left(5 {x}^{2} - 2 {\omega}^{2} {y}^{2}\right)$

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

$= {\left(5 {x}^{2}\right)}^{2} - 2 \left(- 1\right) 5 {x}^{2} \cdot 2 {y}^{2} + 1 \cdot {\left(2 {y}^{2}\right)}^{2}$

$= 25 {x}^{4} + 10 {x}^{2} {y}^{2} + 4 {y}^{4}$

So there are quadratic factors, but the ones we have found have complex coefficients.

It is possible to factor these further, into linear factors with complex coefficients with more square roots, but I think you get the idea that there are no simpler factors with real coefficients.