$ \dfrac {2}{3}x^{-\dfrac {1}{3}} $
So $(2/3)x^{(- 1/3)}$
How to write this in a fraction using roots?
$ \dfrac {2}{3}x^{-\dfrac {1}{3}} $
So $(2/3)x^{(- 1/3)}$
How to write this in a fraction using roots?
$\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3}\cdot\frac{1}{x^{\frac{1}{3}}}=\frac{2}{3}\cdot\frac{1}{\sqrt[3] x}=\frac{2}{3}\cdot\frac{1}{\sqrt[3] x}\cdot\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}=\frac{2\sqrt[3]{x^2}}{3\sqrt[3]{x^3}}=\frac{2\sqrt[3]{x^2}}{3|x|}$
or
$\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3}\cdot\frac{1}{x^{\frac{1}{3}}}=\frac{2}{3}\cdot\frac{1}{\sqrt[3] x}=\frac{2}{3}\cdot\frac{1}{\sqrt[3] x}\cdot\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}=\frac{2\sqrt[3]{x^2}}{3\sqrt[3]{x^3}}=\frac{2}{3}\sqrt[3]{\frac{x^2}{x^3}}=\frac{2}{3}\sqrt[3]{\frac{1}{x}}$
We have implement the formula:
1) $a^{-n}=\frac{1}{a^n}$
2) $a^{\frac{m}{n}}=\sqrt[n]{x^m}$
There are many possibilities. One is $\sqrt[3]{\dfrac{8}{27x}}$
Remember that $a^{-n} = \frac{1}{a^n},$ so $ \frac{2}{3}x^{-\frac{1}{3}} = \frac{2}{3x^{\frac{1}{3}}} = \frac{2}{3\sqrt[3]{x}}. $
$ \frac{2}{3}x^{-\frac{1}{3}} = \frac{2}{3} \cdot \frac{1}{x^{\frac{1}{3}}} = \frac{2}{3}\frac{1}{\sqrt[3]{x}} = \frac{2}{3\sqrt[3]{x}}. $