<s>[INST] Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:

Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.

I struggle to even start this question. 
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ as a solution, though I do not know how to prove that there are no other solutions. 
Upon differentiation, I obtain:
\begin{align}
\frac{d}{dx} (\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4})&= \frac{1}{3}(\sqrt[3]{(x+2)^{-2}}+\sqrt[3]{(x+3)^{-2}}+\sqrt[3]{({x+4})^{-2}})\\
\end{align}
which is always positive for all real values of $x$, implying that the function defined as $f(x)=\sqrt[3]{x+2} + \sqrt[3]{x+3} + \sqrt[3]{x+4}$ is strictly increasing. 
Is there a better way to solve this equation?
 [/INST] $y:=x+3$;
$f(y)=(y-1)^{1/3}+y^{1/3}+(y+1)^{1/3}=0$;
By inspection $y=0$ is a solution.
0) $y^{1/3}$ is an odd, increasing function.
1) $y \ge 1$ is ruled out, since $f(y)>0$
2) $y \le -1$ is ruled out, since $f(y)<0$.
Remains $1< y <1$;
3) $f(y)> 0$ for $0<y<1$; since  $-1< (y-1) <0$,  $1<(y+1) <2$: 
4) Similarly $f(y) <0$ for $-1<y <0$;
It follows $y=0$ is the only solution.
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