Answer
$\color{blue}{\dfrac{m\sqrt[3]{n^2}}{n}}$
Work Step by Step
RECALL:
(1) For nonnegative real numbers $a$ and $b$, $\sqrt[n]{a}\cdot \sqrt{b} = \sqrt[n]{ab}$.
(2) For nonnegative odd integers, $\sqrt[n]{a^n} = a$
(3) $a^m\cdot a^n=a^{m+n}$
Use rule s(1) and (3) above to obtain:
$=\dfrac{\sqrt[3]{m^{1+2}n}}{\sqrt[3]{n^2}}
\\=\dfrac{\sqrt[3]{m^3n}}{\sqrt[3]{n^2}}$
Rationalize the denominator by multiplying $\sqrt[3]{n}$ to both the numerator and the denominator to obtain:
$=\dfrac{\sqrt[3]{m^3n(n)}}{\sqrt[3]{n^2(n)}}
\\=\dfrac{\sqrt[3]{m^3n^2}}{\sqrt[3]{n^3}}
\\=\dfrac{\sqrt[3]{m^3n^2}}{n}$
Simplify the numerator by bringing out of the radical sign the cube root of $m^3$ to obtain:
$=\color{blue}{\dfrac{m\sqrt[3]{n^2}}{n}}$
