Answer     $1$

Let $\alpha = (2 + \sqrt{5})^{1/3} - (-2 + \sqrt{5})^{1/3}$. Then by raising both sides to the third power, expanding and simplying, we have $$\alpha^3 + 3\alpha-4=0$$

This means that $\alpha$ is one of the rational roots to the equation $$x^3+3x-4=0\Leftrightarrow (x-1)(x^2+x+4)=0$$

However, $x^2+x+4=0$ has no real root. This means that $\alpha$ must be the root of $x-1=0$, i.e, $\alpha=1$.