Is $\sqrt[3]{-3} = -\sqrt[3]{3}$ true? Well, Wolfram Alpha says no. Why?
$$\sqrt[3]{-3} = -\sqrt[3]{3} \iff (\sqrt[3]{-3})^3 = (-1)^3(\sqrt[3]{3})^3 \iff -3=(-1)3 \iff -3=-3$$
So I would say it is true if we are talking about real numbers.
In fact in &quot;Analisi Matematica&quot; 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 $\alpha &gt;0$, there exists one, and only one, real number $\beta&gt;0$ such that
$$\beta^n=\alpha \mbox.$$
Definition 1.7.3 The number $\beta &gt;0$  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 $\alpha &gt;0$. We set
$$\alpha^{m/n}=(\sqrt[n]\alpha )^m $$
Observation: we now want to define all of this for negative real numbers. Let $-a&lt;0$ 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 $\beta &gt;0$ be such that $\beta^{2k+1}=\alpha$ ($&gt;0$), 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 $a&gt;0$), showing that the following definition is well-defined:
Definition 1.7.5:
(1) if $\alpha \ge 1$ and $\beta &gt;0$ we set
$$\alpha^{\beta}:= \sup\biggl{\{} \alpha^{m/n}:\frac{m}{n} \le \beta \biggr{\}}$$
(2) if $0&lt;\alpha &lt; 1$ and $\beta &gt;0$ we set
$$\alpha^{\beta}:=\frac{1}{(1/\alpha)^{\beta}}$$
(3) if $\alpha &gt;0$ and $\beta &gt;0$ we set
$$\alpha^{-\beta}=\frac{1}{\alpha^{\beta}$$
(4) $\alpha^0=1$
(defined for every $\alpha&gt;0$ and $\beta$ reals).
$$\sqrt[3]{-3} = -\sqrt[3]{3} \iff (\sqrt[3]{-3})^3 = (-1)^3(\sqrt[3]{3})^3 \iff -3=(-1)3 \iff -3=-3$$
So I would say it is true if we are talking about real numbers.
In fact in &quot;Analisi Matematica&quot; 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 $\alpha &gt;0$, there exists one, and only one, real number $\beta&gt;0$ such that
$$\beta^n=\alpha \mbox.$$
Definition 1.7.3 The number $\beta &gt;0$  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 $\alpha &gt;0$. We set
$$\alpha^{m/n}=(\sqrt[n]\alpha )^m $$
Observation: we now want to define all of this for negative real numbers. Let $-a&lt;0$ 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 $\beta &gt;0$ be such that $\beta^{2k+1}=\alpha$ ($&gt;0$), 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 $a&gt;0$), showing that the following definition is well-defined:
Definition 1.7.5:
(1) if $\alpha \ge 1$ and $\beta &gt;0$ we set
$$\alpha^{\beta}:= \sup\biggl{\{} \alpha^{m/n}:\frac{m}{n} \le \beta \biggr{\}}$$
(2) if $0&lt;\alpha &lt; 1$ and $\beta &gt;0$ we set
$$\alpha^{\beta}:=\frac{1}{(1/\alpha)^{\beta}}$$
(3) if $\alpha &gt;0$ and $\beta &gt;0$ we set
$$\alpha^{-\beta}=\frac{1}{\alpha^{\beta}$$
(4) $\alpha^0=1$
(defined for every $\alpha&gt;0$ and $\beta$ reals).