Answer
$2x^3y^5\sqrt[3]{4y^2}$
Work Step by Step
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:
$=\sqrt[3]{(2^3x^9y^{15})\cdot 2^2y^2}
\\=\sqrt[3]{2^3(x^3)^3(y^5)^3 \cdot 4y^2}
\\=2x^3y^5\sqrt[3]{4y^2}$
