<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 $\sqrt[3]{x+2}=a$ etc.
So, $a+b+c=0$
and  $a^3+c^3=2b^3=2(-a-c)^3=-2(a^3+c^3+3a^2c+3ac^2)$
$$\iff0=a^3+ c^3+2a^2c+2ac^2=(a+c)(a^2-ac+c^2)+2ac(c+a)=(c+a)(a^2+ca+c^2)$$
But $a^2+ca+c^2=0$
$\implies$ either $a=c=0$
or $\dfrac ac=$ imaginary as $\left(\dfrac ac\right)^2+\dfrac ac+1=0$
$\implies a+c=0\implies a^3+c^3=0$
</s>