Question 677000
{{{x^3y^6-27}}}


{{{(xy^2)^3-3^3}}}


{{{(xy^2-3)((xy^2)^2+3xy^2+(3)^2)}}} Use the difference of cubes factoring rule here.


{{{(xy^2-3)(x^2y^4+3xy^2+9)}}}


So {{{x^3y^6-27}}} factors to {{{(xy^2-3)(x^2y^4+3xy^2+9)}}}