<s>[INST] When is the degree of this number 3? I am helping a friend of mine, that works in history of mathematics. She is studying the story of the solution of the cubic equation by Cardano. Sometimes she asks me some mathematical questions, that are very hard to motivate from a modern point of view, but that were interesting to Cardano. So please do not ask for motivations. The question is the following. Let $a$, $b$ be rational numbers, with $b$ not a square. Consider the number
$$
t=\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}-\sqrt{a^2-b}
$$
Under what conditions on $a$ and $b$ is the degree (over $\mathbb{Q}$) of $t$ equal to $3$? A sufficient condition can be found as follows. Let $P(x) = x^3+\alpha_2 x^2 + \alpha_1 x + \alpha_0$ be a rational polynomial. The general expression of the roots of $P$ is
$$
\sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \frac{\alpha_2}{3},
$$
where
$$
q = \frac{2\alpha_2^3 - 9\alpha_2\alpha_1 + 27\alpha_0}{27}
$$
and
$$
p = \frac{3\alpha_1 - \alpha_2^2}{3},
$$
see here. So we can take $a = -\frac{q}{2}$ and $b = \frac{q^2}{4} + \frac{p^3}{27}$ and we need to force $\sqrt{a^2 -b} = \frac{\alpha_2}{3}$. This boils down to $\alpha_1 = \frac{\alpha_2^2 - 3 \sqrt[3]{3\alpha_2^2}}{3}$. We find that one of the solutions of
$$
x^3 + \alpha_2 x^2+ \frac{\alpha_2^2 - 3 \sqrt[3]{3\alpha_2^2}}{3}x+\alpha_0=0
$$
has the required form (of course we need to assume that $\alpha_2$ is such that $\sqrt[3]{3 \alpha_2^2}$ is rational). In this case $a$ and $b$ are given by the above expressions.
I suspect that if $t$ has degree $3$, then its minimal polynomial must be of this form and that $a$ and $b$ are as above, but I am not able to prove it. Note that the condition that $t$ has degree $3$ implies that it can be written as the sum of two cubic root and a rational number (because of the formula), but it is not completely clear that this way of writing $t$ is unique.
 [/INST] $$t:= \sqrt[3]{d(1 + x^6) + \sqrt{d^2 (1 - x^6)^2}} + \sqrt[3]{d(1 + x^6) - \sqrt{d^2 (1 - x^6)^2}} 
 + \sqrt{d^2(1 +x^6)^2 - d^2(1 - x^6)^2}$$
$$= \sqrt[3]{2 d} + \sqrt[3]{2 d x^6} + \sqrt{4 d^2 x^6}$$
$$
= (1 + x^3) \sqrt[3]{2d} + 2 d x^3;$$
$$\frac{p^3}{27} = \left(\frac{q^2}{4} +  \frac{p^3}{27}\right) - \left(\frac{q}{2}\right)^2 =^{?} b - a^2 = d^2 (1 - x^6)^2 - d^2(1 + x^6)^2 = - 4d^2 x^6 
\Rightarrow 2d \in \mathbf{Q}^3$$
</s>