Answer
$\dfrac{3\sqrt[3]{m^2n}}{m^2n^3}$
Work Step by Step
Rationalizing the denominator of $
\sqrt[3]{\dfrac{27}{m^4n^8}}
$ results to
\begin{array}{l}
\sqrt[3]{\dfrac{27}{m^3n^6}\cdot\dfrac{1}{mn^2}}
\\\\=
\dfrac{3}{mn^2}\sqrt[3]{\dfrac{1}{mn^2}}
\\\\=
\dfrac{3}{mn^2}\sqrt[3]{\dfrac{1}{mn^2}\cdot\dfrac{m^2n}{m^2n}}
\\\\=
\dfrac{3}{mn^2}\sqrt[3]{\dfrac{m^2n}{m^3n^3}}
\\\\=
\dfrac{3\sqrt[3]{m^2n}}{mn^2\cdot mn}
\\\\=
\dfrac{3\sqrt[3]{m^2n}}{m^2n^3}
.\end{array}
