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 $$x^3+3x-4=0 \\(x-1)(x^2+x+4)\to\\ x=1,x^2+x+4=0 ,\Delta <0\\x=1$$
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 $$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=\tfrac12+\tfrac{\sqrt 5}{2}+\tfrac12-\tfrac{\sqrt 5}2=1$$
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.
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.
If you don't have to prove that $x \in \mathbb Z$, then it's easy:
$\hphantom-1 < 2+\sqrt{5} < 5 \implies \hphantom-1 < \sqrt[3]{2+\sqrt{5}} < 2$
$-1 < 2-\sqrt{5} < 0 \implies -1 < \sqrt[3]{2+\sqrt{5}} < 0$
Therefore, $ 0 < x < 2 $ and so $x=1$.
If you know the cubic formula for solving $$x^3 + px + q = 0$$ that helps. The formula is $$x = \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$ Given the "$2$"s in your expression, you might be led to consider $q=-4$, so the cubic $x^3 + px -4 = 0$ with solution $$x = \sqrt[3]{2+\sqrt{4+\frac{p^3}{27}}}+\sqrt[3]{2-\sqrt{4+\frac{p^3}{27}}}$$ Now it's clear that $p=3$ gives your expression. So your expression is a solution to $$x^3+3x-4=0$$ which factors as $$(x-1)(x^2+x+4)=0$$ The quadratic factor has no real roots. So your number must be $1$.
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. $$\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}=\frac{1}{2}\left(\sqrt[3]{16+8\sqrt5}+\sqrt[3]{16-8\sqrt5}\right)=$$ $$=\frac{1}{2}\left(\sqrt[3]{1+3\sqrt5+3\cdot5+5\sqrt5}+\sqrt[3]{1-3\sqrt5+3\cdot5-5\sqrt5}\right)=$$ $$=\frac{1}{2}\left(1+\sqrt5+1-\sqrt5\right)=1.$$