Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$ \dfrac {2}{3}x^{-\dfrac {1}{3}} $

So $(2/3)x^{(- 1/3)}$

How to write this in a fraction using roots?

share|improve this question
add comment

4 Answers 4

up vote 2 down vote accepted

Remember that $$a^{-n} = \frac{1}{a^n},$$ so $$ \frac{2}{3}x^{-\frac{1}{3}} = \frac{2}{3x^{\frac{1}{3}}} = \frac{2}{3\sqrt[3]{x}}. $$

share|improve this answer
add comment

There are many possibilities. One is $$\sqrt[3]{\dfrac{8}{27x}}$$

share|improve this answer
add comment

$$ \frac{2}{3}x^{-\frac{1}{3}} = \frac{2}{3} \cdot \frac{1}{x^{\frac{1}{3}}} = \frac{2}{3}\frac{1}{\sqrt[3]{x}} = \frac{2}{3\sqrt[3]{x}}. $$

share|improve this answer
Thank you, that was my answer too however the correction model is wrong for the millionth time, that's why I had to ask it! –  ZafarS Oct 2 '12 at 16:59
add comment

$\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3}\cdot\frac{1}{x^{\frac{1}{3}}}=\frac{2}{3}\cdot\frac{1}{\sqrt[3] x}=\frac{2}{3}\cdot\frac{1}{\sqrt[3] x}\cdot\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}=\frac{2\sqrt[3]{x^2}}{3\sqrt[3]{x^3}}=\frac{2\sqrt[3]{x^2}}{3|x|}$


$\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3}\cdot\frac{1}{x^{\frac{1}{3}}}=\frac{2}{3}\cdot\frac{1}{\sqrt[3] x}=\frac{2}{3}\cdot\frac{1}{\sqrt[3] x}\cdot\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}=\frac{2\sqrt[3]{x^2}}{3\sqrt[3]{x^3}}=\frac{2}{3}\sqrt[3]{\frac{x^2}{x^3}}=\frac{2}{3}\sqrt[3]{\frac{1}{x}}$

We have implement the formula:

1) $a^{-n}=\frac{1}{a^n}$

2) $a^{\frac{m}{n}}=\sqrt[n]{x^m}$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.