$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$ is rational or irrational? 
The number $x$ defined below is rational or irrational?
  $$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$$
  From: IMO 1973 - Longlist

My attempt (my real question is at the end): 
the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab-ac-bc)$ when $a+b+c=0$, leads to $$a^3+b^3+c^3=3abc \tag{1}$$ 
Now considering
$$a=\sqrt[3]{\sqrt{5}+2},b=\sqrt[3]{\sqrt{5}-2},c=-x$$
from (1) it is true that 
$$x^3-3x-2\sqrt{5}=0 \tag{2}$$
That is the number $x$ is a root from (2). 
Note: By trial and error I've found that answer is $x=\sqrt{5}$ (the other 2 roots are complex), that is irrational. But my question is more subtle.
Question: Can I conclude just inspecting (2), judging by the coefficient $2\sqrt{5}$, that $x$ is irrational, without actually solving the equation? In a math contest that might be helpful, if possible, as it would avoid extra steps. 
 A: Alternatively
$$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}=\\
\sqrt[3]{\frac{8\sqrt{5}+16}{8}}+\sqrt[3]{\frac{8\sqrt{5}-16}{8}}=\\
\sqrt[3]{\frac{5\sqrt{5}+3\cdot5+3\sqrt{5}+1}{8}}+\sqrt[3]{\frac{5\sqrt{5}-3\cdot5+3\sqrt{5}-1}{8}}=\\
\sqrt[3]{\frac{\left(\sqrt{5}+1\right)^3}{2^3}}+\sqrt[3]{\frac{\left(\sqrt{5}-1\right)^3}{2^3}}=\\
\frac{\sqrt{5}+1}{2}+\frac{\sqrt{5}-1}{2}=\\
\sqrt{5}$$ 
is irrational.
A: Let $\alpha=\sqrt[3]{\sqrt{5}+2}$ and $\beta=\sqrt[3]{\sqrt{5}-2}$. Then $\alpha\beta=1$ and 
$$ 2\sqrt{5} = \alpha^3+\beta^3 = (\alpha+\beta)((\alpha+\beta)^2-3\alpha\beta)=(\alpha+\beta)^3-3(\alpha+\beta),$$
so the rationality of $\alpha+\beta$ would imply the rationality of $2\sqrt{5}$. It follows that $\alpha+\beta\not\in\mathbb{Q}$.
A: Yes you can.
Notice that we have :
$$x^3-3x -2\sqrt{5} = 0 \Leftrightarrow x(x^2-3) = 2\sqrt{5}$$
Hence if $x$ is rational, $x^2-3$ is rational so $x(x^2-3)$ is rational.
Hence because $2\sqrt{5}$ is irrational, that’s a contradiction and $x$ is not rational.
A: If $a=\sqrt[3]{\sqrt{5}+2},b=\sqrt[3]{\sqrt{5}-2}$
Clearly, $a-b>0$
$ab=1$ and $a^3-b^3=4$
$$(a-b)^3+3ab(a-b)=4$$
$$\iff(a-b)^3+3(a-b)-4=0$$
Observe that only positive real root of $t^3+3t-4=0$ is $1$
$\implies a-b=1$
$\implies a+b=+\sqrt{(a-b)^2+4ab}=?$
