Answer
$\dfrac{\sqrt[3]{4}}{2}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\dfrac{\sqrt[3]{8m^2n^3}\cdot\sqrt[3]{2m^2}}{\sqrt[3]{32m^4n^3}}
,$ use the laws of radicals and the laws of exponents. Rationalize the denominator when necessary.
$\bf{\text{Solution Details:}}$
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel}
\dfrac{\sqrt[3]{8m^2n^3(2m^2)}}{\sqrt[3]{32m^4n^3}}
.\end{array}
Using the Quotient Rule of radicals which is given by $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}{},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\sqrt[3]{\dfrac{8m^2n^3(2m^2)}{32m^4n^3}}
.\end{array}
Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\sqrt[3]{\dfrac{16m^{2+2}n^3}{32m^4n^3}}
\\\\=
\sqrt[3]{\dfrac{16m^{4}n^3}{32m^4n^3}}
\\\\=
\sqrt[3]{\dfrac{\cancel{16}\cancel{m^{4}n^3}}{\cancel{32}^2\cancel{m^{4}n^3}}}
\\\\=
\sqrt[3]{\dfrac{1}{2}}
.\end{array}
Rationalizing the denominator results to
\begin{array}{l}\require{cancel}
\sqrt[3]{\dfrac{1}{2}\cdot\dfrac{4}{4}}
\\\\=
\sqrt[3]{\dfrac{1}{8}\cdot4}
\\\\=
\sqrt[3]{\left(\dfrac{1}{2}\right)^3\cdot4}
\\\\=
\dfrac{1}{2}\sqrt[3]{4}
\\\\=
\dfrac{\sqrt[3]{4}}{2}
.\end{array}
