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 $x=1$.

We can use $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$.
Since $a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}((a-b)^2+(a-c)^2+(b-c)^2)$, we see that
$$a^2+b^2+c^2-ab-ac-bc=0\Leftrightarrow a=b=c.$$
Thus, since $\sqrt[3]{2+\sqrt5}\neq\sqrt[3]{2-\sqrt5}$, we obtaion
$$x=\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}$$ it's
$$x-\sqrt[3]{2+\sqrt5}-\sqrt[3]{2-\sqrt5}=0$$ or
$$x^3-2-\sqrt5-2+\sqrt5-3x\sqrt[3]{(2+\sqrt5)(2-\sqrt5)}=0$$ or
$$x^3+3x-4=0$$ or
$$x^3-x^2+x^2-x+4x-4=0$$ or
$$(x-1)(x^2+x+4)=0$$ or
$$x=1.$$
We can make also the following.
$$\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}=\frac{1}{2}\left(\sqrt[3]{16+8\sqrt5}+\sqrt[3]{16-8\sqrt5}\right)=$$
$$=\frac{1}{2}\left(\sqrt[3]{1+3\sqrt5+3\cdot5+5\sqrt5}+\sqrt[3]{1-3\sqrt5+3\cdot5-5\sqrt5}\right)=$$
$$=\frac{1}{2}\left(1+\sqrt5+1-\sqrt5\right)=1.$$
