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

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Answer 1 · 2015-08-29T13:09:36

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

$=(x^2+4y^2-1)(x^4+16y^4+1-4x^2y^2+4y^2+x^2)$

We can at least partially factor this as follows:

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

Then

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

This symmetric polynomial is a little easier to work with:

$u^3+v^3+w^3-3uvw$

$= (u+v+w)(u^2+v^2+w^2-uv-vw-wu)$

$=(x^2+4y^2-1)(x^4+16y^4+1-4x^2y^2+4y^2+x^2)$

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

By the way, $u^2+v^2+w^2-uv-vw-wu$ has linear factors with Complex coefficients:

$u^2+v^2+w^2-uv-vw-wu$

$= (u+omega v + omega^2 w)(u+omega^2 v + omega w)$

where $omega = -1/2+sqrt(3)/2 i$ is the primitive cube root of $1$ .

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