Answer
$\dfrac{3y}{x}$
Work Step by Step
RECALL:
(i) $a^{-m} = \dfrac{1}{a^m},a\ne0$
(ii) $\sqrt[n]{a^m} = a^{\frac{m}{n}}$
(iii) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$
Simplify the radicand by canceling common factors:
$\require{cancel}
=\sqrt[3]{\dfrac{\cancel{54}27\cancel{x^2}\cancel{y^4}y^3}{\cancel{2}\cancel{x^5}x^3\cancel{y}}}
\\=\sqrt[3]{\dfrac{27y^3}{x^3}}$
Factor the radicand so that at least one factor is a perfect cube to obtain:
$\\=\sqrt[3]{\dfrac{3^3y^3}{x^3}}$
Simplify to obtain:
$=\dfrac{3y}{x}$
