We want only the real 3rd root.
By calculation, $[\sqrt[3]{\sqrt 5 +2}-\sqrt[3]{\sqrt 5 -2}]^3= 4-3[\sqrt[3]{\sqrt 5 +2}-\sqrt[3]{\sqrt 5 -2}]$
Therefore, the answer is a root of $t^3=4-3t$ , which obviously has the real solution $t =1$.
But I want another way of showing $\sqrt[3]{\sqrt 5 +2}-\sqrt[3]{\sqrt 5 -2} =1$, maybe by using simple algebraic formulas.
By observation,
$$(\sqrt5+1)^3=16+8\sqrt5=8(\sqrt5+2)$$
$$\implies\left(\dfrac{\sqrt5+1}2\right)^3=\sqrt5+2$$
Similarly,$$\left(\dfrac{\sqrt5-1}2\right)^3=\sqrt5-2$$
Motivation:
$$(\sqrt5+a)^3=a^3+15a+\sqrt5(3a^2+5)$$
Let us find $a$ such that $$\dfrac{a^3+15a}{3a^2+5}=\dfrac21\iff0=a^3-6a^2+15a-10=(a-1)(a^2-5a+10)$$
Observe that the only real root is $a=1$
