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 $x=1$.

By observing that 
$$\left(\frac12\pm\frac{\sqrt 5}{2}\right)^3=\frac18\pm\frac{3\sqrt 5}8+\frac{15}{8}\pm\frac{5\sqrt 5}{8}=2\pm\sqrt 5$$
It follows
$$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=\tfrac12+\tfrac{\sqrt 5}{2}+\tfrac12-\tfrac{\sqrt 5}2=1$$
