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I was taught in school that

$$x^{a/b} = \sqrt[b]{x^a}$$

however, wolfram says this is not always true:

$\sqrt[3]{x^2} \ne x^{2/3}$

but also says:

$\sqrt[3]{x} = x^{1/3}$

Is he right? If so, can anyone explain why?

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On WolframAlpha, it is mentioned that this holds for $x \geq 0$. For negative $x$, $\sqrt[3]{x^2}$ is defined and is positive, while $x^{2/3}$ is taken negative as a principal value (at least there). These functions are best considered as functions of a complex variable though, with multiple branches. – A Walker Oct 19 '11 at 7:03
For example, for $(-1)^{3/2}$ it returns $e^{\frac{2i\pi}{3}}=-\frac12+i\frac{\sqrt3}2$. – Brian M. Scott Oct 19 '11 at 7:11
The problem is your cube root: there are always three (complex) numbers that when cubed yield your original number. Wolfram Alpha takes the "principal value" of the cube root of a negative number to be the cube root with positive imaginary part. – J. M. Oct 19 '11 at 7:19
Thank you, that pretty much explained everything. – rogi Oct 19 '11 at 7:28

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