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

$2(x^{a}+2y^{b})(x^{2a}-2x^ay^{b}+4y^{2b})$
Factoring the $GCF= 2,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 2x^{3a}+16y^{3b} \\\\= 2(x^{3a}+8y^{3b}) .\end{array} The expressions $x^{3a}$ and $8y^{3b}$ are both perfect cubes (the cube root is exact). Hence, $x^{3a}+8y^{3b}$ is a $\text{ sum }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2[(x^{a})^3+(2y^{b})^3] \\\\= 2(x^{a}+2y^{b})[(x^{a})^2-x^a\cdot (2y^{b})+(2y^{b})^2] \\\\= 2(x^{a}+2y^{b})(x^{2a}-2x^ay^{b}+4y^{2b}) .\end{array}