Answer
$\frac{\sqrt [3] {245x^2y}}{7xy}$
Work Step by Step
First, we need to cancel out common terms in the numerator and denominator:
$\sqrt [3] {\frac{5}{7xy^2}}$
Separate the radical:
$\frac{\sqrt [3] {5}}{\sqrt [3] {7xy^2}}$
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 7's, two more x's, and one more y to eliminate the radical. We multiply the numerator by the same factor:
$\frac{\sqrt [3] {5}}{\sqrt [3] {7xy^2}} • \frac{\sqrt [3] {7^2x^2y}}{\sqrt [3] {7^2x^2y}}$
Multiply to simplify:
$\frac{\sqrt [3] {5 • 49 • x^2 • y}}{\sqrt [3] {7^3 • x^3 • y^3}}$
Take the cube root of any perfect cubes:
$\frac{\sqrt [3] {245x^2y}}{7xy}$
