Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$(x^4-yz^4)(x^8+x^4yz^4+y^2z^8)$
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}