Answer
$\color{blue}{\dfrac{\sqrt[3]{4}}{2}}$
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 real numbers $a$ and $b$ where $b\ne0$, $\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac{a}{b}}$.
(3) For nonnegative odd integers, $\sqrt[n]{a^n} = a$
(4) $a^m\cdot a^n=a^{m+n}$
Use rules (1) and (4) above to obtain:
$=\dfrac{\sqrt[3]{16m^{2+2}n^3}}{\sqrt[3]{32m^4n^3}}
\\=\dfrac{\sqrt[3]{16m^4n^3}}{\sqrt[3]{32m^4n^3}}$
Use rule (2) above to obtain:
$\require{cancel}=\sqrt[3]{\dfrac{16m^4n^3}{32m^4n^3}}
\\=\sqrt[3]{\dfrac{\cancel{16}\cancel{m^4n^3}}{\cancel{32}^2\cancel{m^4n^3}}}
\\=\sqrt[3]{\dfrac{1}{2}}$
Rationalize the denominator by multiplying $\sqrt[3]{2^2}$ to both the numerator and the denominator to obtain:
$=\sqrt[3]{\dfrac{1(2^2)}{2(2^2)}}
\\=\sqrt[3]{\dfrac{4}{2^3}}
\\=\color{blue}{\dfrac{\sqrt[3]{4}}{2}}$
