Answer
$-2y^{3}$
Work Step by Step
Apply the rule $(ab)^m=a^mb^m$:
$(-32)^{\frac{1}{5}} \cdot (y^{15})^{\frac{1}{5}}$
Use the rule $(a^m)^n=a^{mn}$:
$(-32)^{\frac{1}{5}} \cdot y^{\frac{15}{5}}$
Simplify exponents:
$(-32)^{\frac{1}{5}} \cdot y^{3}$
Change fractional exponents into radicals:
$\sqrt[5] {-32} \cdot y^{3}$
Rewrite the radicand as a base raised to the fifth power:
$=\sqrt[5]{-32}\cdot y^{3}\\
=\sqrt[5]{(-2)^5}\cdot y^{3}$
Simplify radicals:
$-2y^{3}$
