Question 8435
Not really.  This is a sum of cubes, which turns out to be a binomial times a trinomial:
{{{x^6 + y^6 }}}
{{{(x^2)^3 + (y^2)^3}}}
{{{(x^2 + y^2)(x^4 - x^2y^2 + y^4) }}}


Neither of these factors can be refactored, so this must be the final answer.


R^2 at SCC