Answer
$\dfrac{2x^2\sqrt[5]{2x^4}}{y^3}$
Work Step by Step
RECALL:
The quotient rule:
$\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$
where
$\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers and $b\ne0$
Use the quotient rule above to obtain:
$=\dfrac{\sqrt[5]{64x^{14}}}{\sqrt[5]{y^{15}}}$
Factor each radicand so that at least one factor is a perfect fifth power to obtain:
$=\dfrac{\sqrt[5]{32x^{10}(2x^4)}}{\sqrt[5]{(y^3)^5}}
\\=\dfrac{\sqrt[5]{(2x^2)^5(2x^4)}}{\sqrt[5]{(y^3)^5}}$
Simplify to obtain:
$=\dfrac{2x^2\sqrt[5]{2x^4}}{y^3}$
