<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] 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.
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