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.

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:
$$\sqrt5+\frac{10}{3\sqrt3}$$
