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?
 A: Let $f(x)=\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}$. 
As $g(x)=x^3$ is increasing, the inverse function $h(x)=x^{1/3}$ is also increasing, so $f(x)$, the sum of 3 increasing functions, is also increasing. By the way, $f(x)$ is also continuous.
Inspection shows that $x=-3$ is a  solution. As the function $f(x)$ is both continuous and increasing, $x=-3$ is the only real solution. 
A: $y:=x+3$;
$f(y)=(y-1)^{1/3}+y^{1/3}+(y+1)^{1/3}=0$;
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.
A: 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 $\left(\dfrac ac\right)^2+\dfrac ac+1=0$
$\implies a+c=0\implies a^3+c^3=0$
A: $$\sqrt[3]{x+2}+\sqrt[3]{x+4}=-\sqrt[3]{x+3}$$
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 $y=0$
A: 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 $x=-3$.
