Answer
$\frac{5m\sqrt [3] {4m^2n^2}}{2n}$
Work Step by Step
We don't want to leave radicals in the denominator, so to get rid of radicals in the denominator, we multiply both the numerator and denominator by as many factors that are needed to eliminate the radical. In this case, we need to multiply the denominator by two more 2's, two more m's, and two more n's to eliminate the radical. We multiply the numerator by the same factor:
$\frac{5m^2}{\sqrt [3] {2m^4n}} • \frac{\sqrt [3] {2^2m^2n^2}}{\sqrt [3] {2^2m^2n^2}}$
Multiply to simplify:
$\frac{5m^2\sqrt [3] {4m^2n^2}}{\sqrt [3] {2^3 • m^6 • n^3}}$
Take the cube root of any perfect cubes:
$\frac{5m^2\sqrt [3] {4m^2n^2}}{2m^2n}$
Divide numerator and denominator by any factors common to both:
$\frac{5m\sqrt [3] {4m^2n^2}}{2n}$
