Answer
$\frac{\sqrt[3] {16}}{\sqrt[3] {16}}$
Work Step by Step
Appropriate form of $1$ that can be used as a factor to eliminate the radical in the denominator is $\frac{\sqrt[3] {16}}{\sqrt[3] {16}}$.
The denominator then becomes
$\sqrt[3] {4}\cdot \sqrt[3] {16}=\sqrt[3] {4^{3}}=4$ and the radical is eliminated.
$\dfrac{3m}{\sqrt[3]4}=\dfrac{3m}{\sqrt[3]4}\times \frac{\sqrt[3]{4^2}}{\sqrt[3]{4^2}}=\frac{3m\sqrt[3]{16}}{4}$
