Answer
$(x^2+xy+y^2)(x^2-xy+y^2)$
Work Step by Step
Calculate the difference of squares:
$$x^6-y^6=(x^3)^2-(y^3)^2=(x^3-y^3)(x^3+y^3).$$
Calculate the difference of cubes:
$$\begin{align*}
x^6-y^6&=(x^2)^3-(y^2)^3\\
&=(x^2-y^2)(x^4+x^2y^2+y^4).
\end{align*}$$
We determine the factorization for $x^4+x^2y^2+y^4$ by rewriting $x^6-y^6$ in two ways:
$$\begin{align*}
(x^3-y^3)(x^3+y^3)&=(x^2-y^2)(x^4+x^2y^2+y^4)\\
x^4+x^2y^2+y^4&=\dfrac{(x^3-y^3)(x^3+y^3)}{x^2-y^2}\\
&=\dfrac{(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)}{(x-y)(x+y)}\\
&=(x^2+xy+y^2)(x^2-xy+y^2).
\end{align*}$$
