Answer
$(p^2-q^2)(p^4+p^2q^2+q^4)$
Work Step by Step
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $
p^6-q^6
,$ is
\begin{array}{l}
(p^2-q^2)[ (p^2)^2-(p^2)(-q^2)+(-q^2)^2]
\\\\=
(p^2-q^2)(p^4+p^2q^2+q^4)
.\end{array}