Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ 
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$

Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result that simplifies to an integer; (b) tried to find $a$ and $b$ such that $(a+\sqrt{b})^3=2+\sqrt{5}$ without success. 
The answer stated for the problem in the original source (a local Math Olympiad Constest) is $x=1$.

If you know the cubic formula for solving $$x^3 + px + q = 0$$ that helps. The formula is $$x = \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$ Given the "$2$"s in your expression, you might be led to consider $q=-4$, so the cubic $x^3 + px -4 = 0$ with solution $$x = \sqrt[3]{2+\sqrt{4+\frac{p^3}{27}}}+\sqrt[3]{2-\sqrt{4+\frac{p^3}{27}}}$$ Now it's clear that $p=3$ gives your expression. So your expression is a solution to $$x^3+3x-4=0$$ which factors as $$(x-1)(x^2+x+4)=0$$ The quadratic factor has no real roots. So your number must be $1$.
