# Question #79897

Oct 16, 2016

$\frac{\sqrt[3]{7 x}}{\sqrt[3]{3 y}} = \frac{\sqrt[3]{63 x {y}^{2}}}{3 y}$

#### Explanation:

Assuming "rationalize" refers to rationalizing the denominator , note that $\sqrt[3]{3 y}$ is the value such that ${\left(\sqrt[3]{3 y}\right)}^{3} = 3 y$. We will use that fact to eliminate the root from the denominator.

$\frac{\sqrt[3]{7 x}}{\sqrt[3]{3 y}} = \frac{\sqrt[3]{7 x} \cdot {\left(\sqrt[3]{3 y}\right)}^{2}}{\sqrt[3]{3 y} \cdot {\left(\sqrt[3]{3 y}\right)}^{2}}$

$= \frac{\sqrt[3]{7 x} \cdot \sqrt[3]{{\left(3 y\right)}^{2}}}{\sqrt[3]{3 y}} ^ 3$

$= \frac{\sqrt[3]{7 x \cdot {\left(3 y\right)}^{2}}}{3 y}$

$= \frac{\sqrt[3]{63 x {y}^{2}}}{3 y}$

(Note that we also used the properties ${\left(\sqrt[a]{x}\right)}^{b} = \sqrt[a]{{x}^{b}}$ and $\sqrt[a]{x} \cdot \sqrt[a]{y} = \sqrt[a]{x y}$ during the process)