A set of numbers Problem: Let $E(x)$ be the number defined by the following expression
\begin{equation*}
 E(x)=\sqrt[3]\frac{x^3-3x+(x^2-1)\sqrt{x^2-4}}{2}+\sqrt[3]\frac{x^3-3x-(x^2-1)\sqrt{x^2-4}}{2} 
\end{equation*}
where  $x$ is a real number and 
$\sqrt[3]{Z}$ denotes the real cubic root of the real number $Z$.
Determine the set
$ \{E(n); n\in \mathbb{N}$ and $n > 1\}$ 

Let: $t=E(n)$ for some $n\gt 1$. So: $$t=\sqrt[3]{\frac {n^3-3n+(n^2-1)\sqrt{n^2-4}}{2}}+\sqrt[3]{\frac {n^3-3n-(n^2-1)\sqrt{n^2-4}}{2}}$$ From here: $$t^3=\frac {n^3-3n+(n^2-1)\sqrt{n^2-4}}{2}+\frac {n^3-3n-(n^2-1)\sqrt{n^2-4}}{2}+3t\sqrt[3]{\frac {(n^3-3n+(n^2-1)\sqrt{n^2-4})(n^3-3n-(n^2-1)\sqrt{n^2-4})}{4}}$$ But: $$(n^3-3n+(n^2-1)\sqrt{n^2-4})(n^3-3n-(n^2-1)\sqrt{n^2-4})=(n^3-3n)^2-(n^2-1)^2(n^2-4)=n^6-6n^4+9n^2-(n^6-6n^4+9n^2-4)=4$$ Thus: $$t^3-3t-n^3+3n=0$$ Which means: $$(t-n)(t^2+n^2+tn-3)=0$$ But the equation: $t^2+n^2+tn-3=0$ hasn't real root for $n\gt 2$. And when $n=2$, we have: $$E(2)=2$$ And finally: $E(n)=n$ for every $n\gt 1$
