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 $x=1$.

You can solve these types of problems by denesting each radical and then simplifying the terms that cancel out. In general, we have$$\sqrt[n]{A+B\sqrt[m]C}=a+b\sqrt[m]{C}$$From which you raise both sides to the nth power and expand using Newton's binomial theorem.

In this case, we have$$\sqrt[3]{2\pm\sqrt5}=a\pm b\sqrt5$$
Cubing both sides, and simplifying, we get the following systems of equations$$\begin{cases}a^3+15ab^2=2\\\\3a^2b+5b^3=1\end{cases}$$
Cross multiplying, and dividing both sides by $b^3$, we get a cubic in $\tfrac ab$. With a rational root at $a/b=1$. Therefore, we see that $a=b$ and$$a^3+15a^3=2\iff 16a^3=2\iff a=b=\frac 12$$Therefore, your radical simplifies into$$\sqrt[3]{2\pm\sqrt5}=\frac 12\pm\frac {\sqrt5}2$$And I will leave the rest of the problem for the OP to finish.
