Is x^1/3 the same as cube root(x)?

Question

I needed to evaluate the surjectivity of cuberoot(x) and used Wolfram-Alpha for that. To my surprise, cuberoot(x) is different than x^1/3, as can be seen A and B.

Further digging shows two contradictory school of thoughts: equal not equal

From my memory, x power 1/n is the nth root of x.

However, article B mentions De Moivre's theorem and I don't think I ever encountered it in the undergrad engineering curriculum.

So are they the same? And in which class do you study this theorem?

Answer 1

When you are dealing with real numbers and only with real numbers, things are quite simple, as far as cube roots are concerned: each real number has one and only one cube root. However, when you introduce complex numbers, things change: each complex number (real or not) different from $0$ has three cube roots.

So, if you desing some software that does computations, you have two choices: either you deal only with real numbers or you consider the possibility of dealing, more generally, with complex numbers. The authors of WolframAlpha chose the second option. And, for their choice, if $r<0$, then its cube root is$$\sqrt[3]{-r}\left(\frac12+\frac{\sqrt3}2i\right);$$here, $\sqrt[3]{-r}$ is the real cube root of $-r$.