Find sum of all positive real numbers $x$ such that $\sqrt[3]{2+x} + \sqrt[3]{2-x}$ becomes an integer. 
Find sum of all positive real numbers $x$ such that $\sqrt[3]{2+x} + \sqrt[3]{2-x}$ becomes an integer.  

I thought Euler's identity may help,so letting $a=\sqrt[3]{2+x}, b=\sqrt[3]{2-x},c=-y$ ($y \in \Bbb Z)$ we have : $a+b+c=0$ leading to :$(2+x)+(2-x)+(-y^3)=3\sqrt[3]{2+x} \sqrt[3]{2-x}(-y)$, but at this point I think sum of such $x's$ can't be found.
 A: $f(x)=\sqrt[3]{2+x} + \sqrt[3]{2-x}$
$$f'(x)=\frac{\sqrt[3]{2-x}^2-\sqrt[3]{x+2}^2}{3 \sqrt[3]{2-x}^2 \sqrt[3]{x+2}^2}$$
$f'(x)=0$  at $x=0$ 
as  $f'(x)>0$ for $x<0$ and $f'(x)<0$ for $x>0$ then $x=0$ is a maximum
and $f(0)=2\sqrt[3]{2}$
$f(x)$ can be integer if $f(x)=2$ that is $x=\dfrac{10}{3 \sqrt{3}}$
or if $f(x)=1$ that is $x=\sqrt{5}$
So the sum of positive $x$ such that $f(x)$ is integer is
$\dfrac{10}{3 \sqrt{3}}+\sqrt{5}$
Hope this can be useful
Edit
Without derivatives
proof that
$$\sqrt[3]{2-x}+\sqrt[3]{x+2}\leq 2 \sqrt[3]{2},\;\forall x\in\mathbb{R}\quad(*)$$
$$\sqrt[3]{2-x}\leq  2 \sqrt[3]{2} - \sqrt[3]{x+2}$$
$$\left(\sqrt[3]{2-x}\right)^3\leq  \left(2 \sqrt[3]{2} - \sqrt[3]{x+2}\right)^3$$
$$2-x\leq -x+14-12\sqrt[3]{4}\sqrt[3]{x+2}+6 \sqrt[3]{2} \sqrt[3]{(x+2)^2}$$
$$-12\leq -12\sqrt[3]{4}\sqrt[3]{x+2}+6 \sqrt[3]{2} \sqrt[3]{(x+2)^2}$$
$$\sqrt[3]{2} \sqrt[3]{(x+2)^2}-2\sqrt[3]{4}\sqrt[3]{x+2}+2\geq 0$$
Set $\sqrt[3]{x+2}=w$
$$\sqrt[3]{2} w^2-2\sqrt[3]{4}w+2\geq 0$$
which is verified for all $w\in\mathbb{R}$
and so is the $(*)$
A: Using Power Mean inequality,it's seen that: $\sqrt[3]{2+x}+\sqrt[3]{2-x}\leq 2\sqrt[3]2$.So the positive integer value of $\sqrt[3]{2+x}+\sqrt[3]{2-x}$ can be $1$ or $2$.
After many hours of thinking,I found out that Euler's identity really helps simplifying the equation:
$$y^3-4=3y\sqrt[3]{4-x^2}\Rightarrow$$
$$x^2=4-\frac{(y^3-4)^3}{27y^3}$$
$$y=1\Rightarrow x=\sqrt5$$
$$y=2\Rightarrow x=\frac{10}{3\sqrt3}$$
So,sum of such $x$'s:
$$\sqrt5+\frac{10}{3\sqrt3}$$
