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

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

How to write this in a fraction using roots?

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4 Answers 4

up vote 2 down vote accepted

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

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

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

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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! – JohnPhteven Oct 2 '12 at 16:59

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


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

We have implement the formula:

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

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

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