# How do you rationalize the denominator and simplify root3(9/7)?

Mar 15, 2016

$\frac{\sqrt[3]{441}}{7}$

#### Explanation:

$1$. Since the denominator contains a cube root, we need to multiply the numerator and denominator by a value that will result in the denominator of $\sqrt[3]{\frac{9}{7}}$ to have a perfect cube. Thus, start by multiplying the numerator and denominator by $\sqrt[3]{49}$.

$\sqrt[3]{\frac{9}{7}}$

$= \frac{\sqrt[3]{9}}{\sqrt[3]{7}}$

$= \frac{\sqrt[3]{9}}{\sqrt[3]{7}} \left(\frac{\sqrt[3]{49}}{\sqrt[3]{49}}\right)$

$2$. Simplify.

$= \frac{\sqrt[3]{441}}{\sqrt[3]{343}}$

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \frac{\sqrt[3]{441}}{7} \textcolor{w h i t e}{\frac{a}{a}} |}}}$