Is it possible?

Let $x=\sqrt[3]{2+\sqrt a} + \sqrt[3]{2-\sqrt a}$ . Then, we have:

$\begin{aligned} x^3 & = \left(\sqrt[3]{2+\sqrt a} + \sqrt[3]{2-\sqrt a}\right)^3 \\ & = 2+\sqrt a + 3\sqrt[3]{(2+\sqrt a)^2(2-\sqrt a)} + 3\sqrt[3]{(2+\sqrt a)(2-\sqrt a)^2} + 2-\sqrt a \\ & = 4 + 3\sqrt[3]{(4-a)(2-\sqrt a)} + 3\sqrt[3]{(2+\sqrt a)(4-a)} \\ & = 4 + 3\sqrt[3]{4-a} \left(\sqrt[3]{2-\sqrt a} + \sqrt[3]{2+\sqrt a} \right) \\ \implies x^3 & = 4 + 3\sqrt[3]{4-a}x \end{aligned}$

For $x$ to be an integer, $4-a$ must be a perfect cube. For $a>0$ , we note that the integral values of $\sqrt[3]{4-a} \le 0$ . Let $n= -\sqrt [3]{4-a}$ . Then $n$ is an non-negative integer. Therefore, we have:

$\begin{aligned} x^3 & = 4 + 3\sqrt[3]{4-a}x \\ x^3 - 3\sqrt[3]{4-a}x & = 4 \\ x^3+3nx & = 4 \\ x(x^2+3n) & = 4 \end{aligned}$

Note that the possible values for $x$ are $-4, -2, -1, 1, 2, 4$ and the only solution is when $x=1$ , then $x^2 + 3n = 4$ $\implies n = 1$ $\implies \sqrt[3]{4-a} = -1$ $\implies a = \boxed{5}$ .

Let's assume $N=(2+\sqrt{a})^{1/3} + (2-\sqrt{a})^{1/3}$

Cubing both side we get $N^3=2+\sqrt{a}+2-\sqrt{a}+3.(4-a)^{1/3}((2+\sqrt{a})^{1/3} + (2-\sqrt{a})^{1/3}) \\ => N^3=4+3N(4-a)^{1/3} \\ =>(N^3-4)^{3}=(3N)^3(4-a)$

Now the value of $a$ will be such that there will be solution in $N$ .The $(4-a)$ should be a perfect cube.

First try with $4-a=1 =>a=3$ but this will not give any integral $N$ .Now $4-a=-1 =>a=5$ this will give integral $N$ .

So, the value of $a=5$ .

Note There may exists other values of $a$ which is not included in the solution.Any suggestion how to prove that this is only value of $a$ ???

If you remember this problem, the solution may help. :)