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?

$$\sqrt[3]{x+2}+\sqrt[3]{x+4}=-\sqrt[3]{x+3}$$
Take Cube in both sides
$$2(x+3)-3\sqrt[3]{(x+2)(x+4)}\sqrt[3]{x+3}=-(x+3)$$
Set $\sqrt[3]{x+3}=y\implies x+3=y^3, (x+2)(x+4)=(y^3-1)(y^3+1)$
$$2y^3-3y\sqrt[3]{y^6-1}=-y^3$$
$$y(y^2-\sqrt[3]{y^6-1})=0$$
If $y\ne0$ $$y^2=\sqrt[3]{y^6-1}$$
Take cube in both sides  $$y^6=y^6-1\iff 0=-1$$ So no finite solutions expect $y=0$
