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

1 Answer
Aug 29, 2015

Answer:

#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)#

Explanation:

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#.