# How do you simplify (root4(5))/(4root4(27))?

Dec 19, 2016

$\frac{\sqrt[4]{15}}{12}$

#### Explanation:

$\frac{\sqrt[4]{5}}{4 \cdot \sqrt[4]{27}}$, we first want to remove the radicals from the denominator, we can do this by multiplying by $\frac{\sqrt[\frac{3}{4}]{27}}{\sqrt[\frac{3}{4}]{27}}$ which is equal to $1$.

$\frac{\sqrt[4]{5}}{4 \cdot \sqrt[4]{27}} \cdot {\sqrt[4]{27}}^{3} / {\sqrt[4]{27}}^{3} = \frac{\sqrt[4]{5} \cdot {\sqrt[4]{27}}^{3}}{4 \cdot \sqrt[4]{27} \cdot {\sqrt[4]{27}}^{3}} = \frac{\sqrt[4]{5} \cdot {\sqrt[4]{27}}^{3}}{4 \cdot {\sqrt[4]{27}}^{4}} = \frac{\sqrt[4]{5} \cdot {\sqrt[4]{27}}^{3}}{4 \cdot 27}$

$= \frac{\sqrt[4]{5} \cdot \sqrt[4]{27} \cdot \sqrt[4]{27} \cdot \sqrt[4]{27}}{108} = \frac{\sqrt[4]{5 \cdot {27}^{3}}}{108} = \frac{\sqrt[4]{5 \cdot {3}^{9}}}{108} = \frac{\sqrt[4]{5 \cdot {3}^{4} \cdot {3}^{4} \cdot 3}}{108} = \frac{3 \cdot 3 \sqrt[4]{5 \cdot 3}}{108} = \frac{9 \sqrt[4]{15}}{9 \cdot 12} = \frac{\cancel{9} \sqrt[4]{15}}{\cancel{9} \cdot 12}$

This is as simplified as it gets,

$\frac{\sqrt[4]{15}}{12}$