#### Answer

$\dfrac{x^2\sqrt[3]{50x^2}}{3y^4}$

#### 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:
$=\dfrac{\sqrt[3]{50x^8}}{\sqrt[3]{27y^{12}}}$
Factor each radicand so that at least one factor is a perfect cube to obtain:
$=\dfrac{\sqrt[3]{x^6(50x^2)}}{\sqrt[3]{(3y^4)^3}}
\\=\dfrac{\sqrt[3]{(x^2)^3(50x^2)}}{3y^4}
\\=\dfrac{x^2\sqrt[3]{50x^2}}{3y^4}$