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 $x=1$.

Let $a=\sqrt[3]{2+\sqrt{5}}, b=\sqrt[3]{2-\sqrt{5}}\,$, then:
$$\require{cancel}
a^3+b^3 = 2+\cancel{\sqrt{5}} + 2-\cancel{\sqrt{5}} = 4 \\
ab =  \sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})} = \sqrt[3]{2^2 - 5} = -1
$$
From $\,(a+b)^3 = a^3+b^3+3ab(a+b)\,$ and given that $\,x=a+b\,$ it follows that $\,x^3=4-3x\,$ $\iff 0 = x^3+3x-4=(x-1)(x^2+x+4)\,$, where only the first factor has a real root.
