# How do you simplify 9/root3(36)?

Mar 15, 2016

$\frac{3 \sqrt[3]{6}}{2}$

#### Explanation:

$1$. Since the denominator of the fraction contains a cube root, we need to multiply the numerator and denominator by a value that will result in the denominator of $\frac{9}{\sqrt[3]{36}}$ to have a perfect cube. Thus, start by multiplying the numerator and denominator by $\sqrt[3]{6}$.

$\frac{9}{\sqrt[3]{36}}$

$= \frac{9}{\sqrt[3]{36}} \left(\frac{\sqrt[3]{6}}{\sqrt[3]{6}}\right)$

$2$. Simplify.

$= \frac{9 \sqrt[3]{6}}{\sqrt[3]{36 \cdot 6}}$

$= \frac{9 \sqrt[3]{6}}{\sqrt[3]{216}}$

$= \frac{9 \sqrt[3]{6}}{6}$

$= \frac{{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}}}^{3} \sqrt[3]{6}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} ^ 2$

$3$. Rewrite the fraction.

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \frac{3 \sqrt[3]{6}}{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$