Wolfram alpha, third root of unity

Question

As I learned, the third root of unity is $e^0$, $e^{\frac{2}{3}i\pi}$ and $e^{-\frac{2}{3}i\pi}$. But in WolframAlpha, when I was calculating $x$ from $x^3=e^t+5$ WolframAlpha said to multiply the third root of unity and multiplied $1$, $-(-1)^{\frac{1}{3}}$ and $(-1)^{\frac{2}{3}}$. I can tell that this is actually the cube root of 1 by cubing it. But how did this number pop out? I did some googling but got nothing.

Answer 1

We have $1 = e^0$, $e^{-\frac{2}{3} i \pi} = -(-1)^{1/3}$ and $e^{i \pi \frac{2}{3}} = (-1)^{2/3}$. This is the polar form of a complex number.

In this case, we can explicitly use Eulers formula $$ e^{ix} = \cos(x) + i \sin(x). $$ Setting $x = \pi$ yields $$ e^{i \pi} = \cos(\pi) + i \sin(\pi) = -1 + 0 = - 1. $$ Therefore, for example $$ (-1)^{\frac{2}{3}} = \left(e^{i \pi}\right)^{\frac{2}{3}} = e^{i \pi \frac{2}{3}}. $$

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Note: We also have $e^{-\frac{2}{3} i \pi} = -e^{\frac{1}{3} i \pi}$. Can you show this with Eulers formula?