## Intermediate Algebra for College Students (7th Edition)

$2x^3y^5\sqrt[3]{4y^2}$
Write 32 as $2^5$ to obtain: $=\sqrt[3]{2^5x^9y^{17}}$ Factor the radicand (expression inside the radical sign) so that at least one factor is a perfect cube, and then simplify to obtain: $=\sqrt[3]{(2^3x^9y^{15})\cdot 2^2y^2} \\=\sqrt[3]{2^3(x^3)^3(y^5)^3 \cdot 4y^2} \\=2x^3y^5\sqrt[3]{4y^2}$