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Well, Wolfram Alpha says no. Why?

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thunder
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    It's true if you stick to real numbers (no complex numbers). Wolframalpha will give a true result if you make the assumption of real numbers. If you're in complex numbers, it assumes the "principal root" which may be complex. – Ameet Sharma Dec 03 '20 at 12:53
  • See what Wolfram Alpha takes as the cubic root of $-3$: https://www.wolframalpha.com/input/?i=%5Csqrt%5B3%5D%7B-3%7D . See also what it takes as a minus cubic root of $3$: https://www.wolframalpha.com/input/?i=-%5Csqrt%5B3%5D%7B3%7D . No wonder it claims they are different. As others have said, $-3$ also has a purely real cubic root, which is equal to $-\sqrt[3]{3}$. –  Dec 03 '20 at 13:04
  • @Ameet Dharma what is the principal cube root of $-3$ in $\mathbb{C}$? –  Dec 03 '20 at 13:05
  • You should also note that to use cube root in Wolfram Alpha, you should use Cbrt(...). Sure enough, you can see that $ \sqrt[3]{-3} = -\sqrt[3]{3} $ – talbi Dec 03 '20 at 16:50
  • @SenZen, I think principal root is usually taken to be the root with the smallest nonnegative angle. In the case of positive numbers, the positive real root is taken to be the principal root (angle 0). In the case of cube root of -3 it is $\sqrt[3]{3}e^{i(\frac{\pi}{3})}$ – Ameet Sharma Dec 03 '20 at 16:58

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$$\sqrt[3]{-3} = -\sqrt[3]{3} \iff (\sqrt[3]{-3})^3 = (-1)^3(\sqrt[3]{3})^3 \iff -3=(-1)3 \iff -3=-3$$ So I would say it is true if we are talking about real numbers.

In fact in "Analisi Matematica" of Paolo Maurizio Sardi (CIttàStudi edizioni), we define roots in the following manner.

Theorem 1.7.1 Given an integer $n \ge 1$, and a real number $\alpha >0$, there exists one, and only one, real number $\beta>0$ such that $$\beta^n=\alpha \mbox.$$

Definition 1.7.3 The number $\beta >0$ found in the Theorem above is called $n$-th root of $\alpha$ and is denoted by $$\beta = \sqrt[n]\alpha \mbox.$$

Observation: $(\sqrt[n]\alpha )^m =\sqrt[n]{\alpha^m}$

Definition 1.7.4 Let $m\in \mathbb{Z}$, let $n\ge 1$ be an integer and let $\alpha >0$. We set $$\alpha^{m/n}=(\sqrt[n]\alpha )^m $$

Observation: we now want to define all of this for negative real numbers. Let $-a<0$ and let $n \ge 1$ be an integer:

(1) if $n=2k$ is even, it is clear that there does not exists $\beta\in \mathbb{R}$ such that $$\beta^{2k}=(\beta^2)^k=-\alpha \mbox,$$ so we cant' do this in the reals;

(2)if $n=2k+1$ is odd and let $\beta >0$ be such that $\beta^{2k+1}=\alpha$ ($>0$), then $$(-\beta)^{2k+1}=-\alpha$$ so $-\beta$ is called the $n$-th root of $-\alpha$ an is denoted by $$-\beta=\sqrt[n]{-\alpha} \mbox.$$

You could then continue with defining roots with real exponent (going back to $a>0$), showing that the following definition is well-defined:

Definition 1.7.5:

(1) if $\alpha \ge 1$ and $\beta >0$ we set $$\alpha^{\beta}:= \sup\biggl{\{} \alpha^{m/n}:\frac{m}{n} \le \beta \biggr{\}}$$

(2) if $0<\alpha < 1$ and $\beta >0$ we set $$\alpha^{\beta}:=\frac{1}{(1/\alpha)^{\beta}}$$

(3) if $\alpha >0$ and $\beta >0$ we set $$\alpha^{-\beta}=\frac{1}{\alpha^{\beta}$$

(4) $\alpha^0=1$

(defined for every $\alpha>0$ and $\beta$ reals).

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