# How do you factor by grouping x^6 + 12x^2y^2 +64y^6 - 1?

Aug 29, 2015

${x}^{6} + 12 {x}^{2} {y}^{2} + 64 {y}^{6} - 1$

$= \left({x}^{2} + 4 {y}^{2} - 1\right) \left({x}^{4} + 16 {y}^{4} + 1 - 4 {x}^{2} {y}^{2} + 4 {y}^{2} + {x}^{2}\right)$

#### Explanation:

We can at least partially factor this as follows:

Let $u = {x}^{2}$, $v = 4 {y}^{2}$ and $w = - 1$

Then

${x}^{6} + 12 {x}^{2} {y}^{2} + 64 {y}^{6} - 1 = {u}^{3} + {v}^{3} + {w}^{3} - 3 u v w$

This symmetric polynomial is a little easier to work with:

${u}^{3} + {v}^{3} + {w}^{3} - 3 u v w$

$= \left(u + v + w\right) \left({u}^{2} + {v}^{2} + {w}^{2} - u v - v w - w u\right)$

$= \left({x}^{2} + 4 {y}^{2} - 1\right) \left({x}^{4} + 16 {y}^{4} + 1 - 4 {x}^{2} {y}^{2} + 4 {y}^{2} + {x}^{2}\right)$

You can then reconstruct this answer as factoring by grouping, but that would not explain how you found the splits.

By the way, ${u}^{2} + {v}^{2} + {w}^{2} - u v - v w - w u$ has linear factors with Complex coefficients:

${u}^{2} + {v}^{2} + {w}^{2} - u v - v w - w u$

$= \left(u + \omega v + {\omega}^{2} w\right) \left(u + {\omega}^{2} v + \omega w\right)$

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