## College Algebra (11th Edition)

$8(y-5z^2)(y^2+5yz^2+25z^4)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $8y^3-1000z^6 ,$ factor first the $GCF.$ Then use the factoring of the sum or difference of $2$ cubes. $\bf{\text{Solution Details:}}$ Factoring the $GCF=8,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 8(y^3-125z^6) .\end{array} Both $y^3$ and $125z^6$ are perfect cubes (the cube root is exact.) 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\begin{array}{l}\require{cancel} \\\\=8[y^3-(5z^2)^3] \\\\=8(y-5z^2)(y^2+5yz^2+25z^4) .\end{array}