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 $x=1$.

$$x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\\x^3=2+\sqrt5+2-\sqrt5+3\sqrt[3]{2+\sqrt{5}}\cdot\sqrt[3]{2-\sqrt{5}}(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})\\x^3=4+3\cdot(-1)\cdot(x)$$so $$x^3+3x-4=0 \\(x-1)(x^2+x+4)\to\\ x=1,x^2+x+4=0 ,\Delta <0\\x=1$$
