$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$ is rational or irrational?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $\alpha+\beta\not\in\mathbb{Q}$.

Answer 4

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 $1$

$\implies a-b=1$

$\implies a+b=+\sqrt{(a-b)^2+4ab}=?$