TITLE: Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
QUESTION [7 upvotes]: 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 $x=1$.

REPLY [3 votes]: If you don't have to prove that $x \in \mathbb Z$, then it's easy:
$\hphantom-1 < 2+\sqrt{5} < 5 \implies \hphantom-1 < \sqrt[3]{2+\sqrt{5}} < 2$
$-1 < 2-\sqrt{5} < 0 \implies -1 < \sqrt[3]{2+\sqrt{5}} < 0$
Therefore, $ 0 < x < 2 $ and so $x=1$.