#### Answer

$\dfrac{3\sqrt [3] {m^2n}}{m^2n^3}$

#### Work Step by Step

$\sqrt[3] {\dfrac{27}{m^4n^8}} = \dfrac{\sqrt[3]{27}}{\sqrt[3] {m^4n^8}} = \dfrac{3}{\sqrt[3] {m^4n^8}} = \dfrac{3}{\sqrt[3] {m^4n^8}} \times \dfrac{\sqrt[3] {m^2n}}{\sqrt[3] {m^2n}} = \dfrac{3\sqrt [3] {m^2n}}{\sqrt[3] {m^4n^8} \sqrt [3] {m^2n}} = \dfrac{3\sqrt [3] {m^2n}}{\sqrt[3] {m^4n^8m^2n}} = \dfrac{3\sqrt [3] {m^2n}}{\sqrt[3] {m^{4+2}n^{8+1}}} = \dfrac{3\sqrt [3] {m^2n}}{\sqrt[3] {m^6n^9}} = \dfrac{3\sqrt [3] {m^2n}}{\sqrt[3] {(m^2)^3(n^3)^3}} = \dfrac{3\sqrt [3] {m^2n}}{m^2n^3}$