Why does Wolfram say domain of $y=\sqrt[3]{(x-1)^2(x+2)}$ is $x \in \mathcal{R} : x \ge 2$

Question

Why does Wolfram say domain of $$y=\sqrt[3]{(x-1)^2(x+2)}$$ is $$x \in \mathcal{R} : x \ge 2$$ when I think is actually all real numbers because it's third root? Even google's quick answer is saying that, but they are probably just using Wolfram and showing the result as their own.

Answer 1

One solution to this can be found when you input the equation in wolfram alpha:

It appears that wolfram alpha considers negative values outside of the domain of the principal root. If you click on "Use the real-valued root" you will get the expected answer.

Answer 2

Thr answer of Wolfram is not correct: in fact $\sqrt[3]{x}$ is defined $\forall x \in R$ (also negative or $0$). So in your case $\sqrt[3]{(x-1)^2(x+2)}$ is defined $\forall x \in R$

Answer 3

WA always thinks that $x^{\frac{1}{3}}=\sqrt[3]{x},$ but it's wrong.

By definition, the domain of $x^r$, where $r\in\mathbb Q$, is $(0,+\infty)$.

The domain of $\sqrt[n]{x}$, where $n\in\mathbb N$, $n$ is even and $n\geq2$, is $[0,+\infty)$;

The domain of $\sqrt[n]{x}$, where $n\in\mathbb N$, $n$ is odd and $n\geq3$, is $\mathbb R$.