A cube root equation

Notation. Let $a=\sqrt[3]{x+1}$, $b=\sqrt[3]{x-1}$, $c=\sqrt[3]{2x}$. The equation is $a+b=c$.

Cubing. $(a+b)^3=a^3+b^3+3ab(a+b)=c^3$, i.e.
$$(x+1)+(x-1)+3\sqrt[3]{(x+1)(x-1)}\cdot c=2x,$$
i.e.
$$2x+3\sqrt[3]{(x^2-1)}\cdot\sqrt[3]{2x}=2x.$$
So $3\sqrt[3]{2x(x^2-1)}=0$, hence $2x(x^2-1)=0$, so $x=0$ or $x=\pm1$.

Check.
- $x=0$: $\sqrt[3]1+\sqrt[3]{-1}=1-1=0=\sqrt[3]0$. OK.
- $x=1$: $\sqrt[3]2+0=\sqrt[3]2$. OK.
- $x=-1$: $0+\sqrt[3]{-2}=-\sqrt[3]2=\sqrt[3]{-2}$. OK.

Conclusion. $x\in\{-1,0,1\}$.