<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] let $t=x+3$ and work with the 'symmetric' form,
$$\sqrt[3]{t-1}+\sqrt[3]{t+1}=-\sqrt[3]{t}\tag{1}$$
Take the cubic power,
$$2t + 3(t^2-1)^{1/3}(\sqrt[3]{t-1}+\sqrt[3]{t+1})=-t$$
Simplify with (1),
$$(t^2-1)t=t^3$$
which leads to the only solution $t=0$, hence $x=-3$.
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