## Intermediate Algebra (6th Edition)

$2(5y-2x)(25y^2+10xy+4x^2)$
Factoring the $GCF= 2$ results to $2(125y^3-8x^3)$. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$, or the factoring of the sum/difference of two cubes, then, \begin{array}{l} 2(125y^3-8x^3) \\= 2[(5y)+(-2x)][(5y)^2-(5y)(-2x)+(-2x)^2) \\= 2(5y-2x)(25y^2+10xy+4x^2) .\end{array}