Answer
$(x^4-yz^4)(x^8+x^4yz^4+y^2z^8)$
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, $
x^{12}-y^3z^{12}
,$ is
\begin{array}{l}
(x^4-yz^4)[ (x^4)^2-(x^4)(-yz^4)+(-yz^4)^2]
\\\\=
(x^4-yz^4)(x^8+x^4yz^4+y^2z^8)
.\end{array}