# How do you simplify root3(4/9)?

Mar 29, 2017

color(blue)(root3(4)/root3(9) or $\textcolor{b l u e}{0.0763142828}$

#### Explanation:

root3(4/(9)

:.=color(blue)(root3(4)/root3(9)

$\therefore = \textcolor{b l u e}{\sqrt[3]{4}} = 1.587401052$

$\therefore \textcolor{b l u e}{\sqrt[3]{9}} = 2.080083823$

$\therefore \frac{1.587401052}{2.080083823}$

$\therefore \textcolor{b l u e}{0.0763142828}$

Mar 29, 2017

${\left(\frac{2}{3}\right)}^{\frac{2}{3}}$

#### Explanation:

$\sqrt[3]{\frac{4}{9}} = \sqrt[3]{{\left(\frac{2}{3}\right)}^{2}}$

$= {\left({\left(\frac{2}{3}\right)}^{2}\right)}^{\frac{1}{3}} = {\left(\frac{2}{3}\right)}^{\frac{2}{3}}$

Mar 29, 2017

$\frac{\sqrt[3]{12}}{3}$
We know that $\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$. The expression can then be simplified to $\sqrt[3]{\frac{4}{9}} = \frac{\sqrt[3]{4}}{\sqrt[3]{9}}$. However, the denominator is an irrational number.
In order to rationalize the denominator, we have to multiply both the numerator and denominator by some factor that makes it rational. The denominator can be expressed as ${9}^{\frac{1}{3}}$. If we multiply it by ${9}^{\frac{2}{3}}$ ($= \sqrt[3]{81} = 3 \sqrt[3]{3}$, this will be used later on), then it becomes rational as ${9}^{\frac{1}{3}} \cdot {9}^{\frac{2}{3}} = 9$.
The fraction then becomes $\frac{\sqrt[3]{4}}{\sqrt[3]{9}} = \frac{\sqrt[3]{4}}{\sqrt[3]{9}} {9}^{\frac{2}{3}} / {9}^{\frac{2}{3}} = \frac{\sqrt[3]{4} \cdot 3 \sqrt[3]{3}}{9} = \frac{\sqrt[3]{12}}{3}$. This is in simplest form.