Answer
$2x^2y^2\sqrt[4]{y^2}$
Work Step by Step
RECALL:
(1) The quotient rule:
$\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\dfrac{a}{b}}$
where
$\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers and $b\ne0$
(2) $\dfrac{a^m}{a^n} = a^{m-n}, a \ne =0$
Use the quotient rule above to obtain:
$\require{cancel}=\sqrt[4]{\dfrac{32x^{10}y^{8}}{2x^2y^{-2}}}
\\=\sqrt[4]{\dfrac{16\cancel{32}\cancel{x^{10}}x^8y^8}{\cancel{2x^2}y^{-2}}}
\\=\sqrt[4]{\dfrac{16x^8y^8}{y^{-2}}}$
Use rule (2) above to obtain:
$=\sqrt[4]{16x^8y^{8-(-2)}}
\\=\sqrt[4]{16x^8y^{8+2}}
\\=\sqrt[4]{16x^8y^{10}}$
Factor the radicand so that at least one factor is a perfect fourth power to obtain:
$=\sqrt[4]{16x^8y^8(y^2)}
\\=\sqrt[4]{2^4(x^2)^4(y^2)^4(y^2)}
\\=\sqrt[4]{(2x^2y^2)^4(y^2)}$
Simplify to obtain:
$=2x^2y^2\sqrt[4]{y^2}$
