# How do you simplify root3(1/7)?

Aug 15, 2017

$\frac{\sqrt[3]{49}}{7}$

#### Explanation:

Since ${x}^{\frac{m}{n}} = \sqrt[n]{{x}^{m}}$, we can write $\sqrt[3]{\frac{1}{7}}$ as ${\left(\frac{1}{7}\right)}^{\frac{1}{3}}$.

According to the power of a quotient rule, ${\left(\frac{a}{b}\right)}^{m} = {a}^{m} / {b}^{m}$. Thus, we can say ${\left(\frac{1}{7}\right)}^{\frac{1}{3}} = {1}^{\frac{1}{3}} / {7}^{\frac{1}{3}}$.

From here, we can say ${1}^{\frac{1}{3}} / {7}^{\frac{1}{3}} = \frac{1}{7} ^ \left(\frac{1}{3}\right)$, since $1$ raised to any power is $1$.

We are left with $\frac{1}{7} ^ \left(\frac{1}{3}\right)$ or $\frac{1}{\sqrt[3]{7}}$.

However, we cannot have a radical in the denominator. To rationalize this expression, we must try to make the denominator $7$. To do this, multiply both the numerator and denominator by ${7}^{\frac{2}{3}} / {7}^{\frac{2}{3}}$ or $\frac{\sqrt[3]{{7}^{2}}}{\sqrt[3]{{7}^{2}}}$, respectively.

$\frac{1}{7} ^ \left(\frac{1}{3}\right) \cdot \textcolor{b l u e}{{7}^{\frac{2}{3}} / {7}^{\frac{2}{3}}} = {7}^{\frac{2}{3}} / {7}^{\frac{3}{3}} = {7}^{\frac{2}{3}} / 7 = \frac{\sqrt[3]{{7}^{2}}}{7} = \frac{\sqrt[3]{49}}{7}$

$\frac{1}{\sqrt[3]{7}} \cdot \textcolor{b l u e}{\frac{\sqrt[3]{{7}^{2}}}{\sqrt[3]{{7}^{2}}}} = \frac{\sqrt[3]{49}}{\sqrt[3]{343}} = \frac{\sqrt[3]{49}}{7}$

So, ${\left(\frac{1}{7}\right)}^{\frac{1}{3}}$ simplifies to $\frac{\sqrt[3]{49}}{7}$.