Question 1122917
Factor the sum 
{{{X^3y^3-64}}}
The special factoring formula for the difference of cubes, namely
{{{(x^3-y^3)=(x-y)(x^2+xy+y^2)}}}
:
{{{X^3y^3-64}}} = {{{((xy)^3-64)}}} = {{{(xy - 4)((xy)^2 + 4xy + 16)}}}