<s>[INST] $x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$ 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. 
 [/INST] 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 $1$
$\implies a-b=1$
$\implies a+b=+\sqrt{(a-b)^2+4ab}=?$
</s>