Answer
$\dfrac{3x^2\sqrt[3]{3x^2}}{2y^5}$
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:
$=\dfrac{\sqrt[3]{81x^8}}{\sqrt[3]{8y^{15}}}$
Factor each radicand so that at least one factor is a perfect cube to obtain:
$=\dfrac{\sqrt[3]{27x^6(3x^2)}}{\sqrt[3]{(2y^5)^3}}
\\=\dfrac{\sqrt[3]{(3x^2)^3(3x^2)}}{\sqrt[3]{(2y^5)^3}}$
Simplify each radical to obtain;
$=\dfrac{3x^2\sqrt[3]{3x^2}}{2y^5}$
