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Recall that a real number $\alpha$ is solvable by real radicals over $\mathbb Q$ iff there is an increasing sequence ${\mathbb Q}= L_0 \subseteq L_1 \subseteq \ldots \subseteq L_r={\mathbb Q}(\alpha)$, such that for each $i$, $L_{i+1}$ is a radical extension of $L_i$, i.e. $L_{i+1}=L_i(\alpha_i^{n_i})$ for some integer $n_i \geq 1$ and some $\alpha_i \in L_i$.

The famous Casus irreducibilis result says that if a cubic has only real and non-rational roots, then none of those roots are solvable by real radicals.

But what about the case when there is only real root, for example for $X^3-(X+1)$ ? Is it known whether the real root is solvable by real radicals or not ?

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  • $\begingroup$ Since the discriminant is negative in this case and the formula contains the third root of the negative of this discriminant, the answer should be yes. $\endgroup$ – Peter May 8 '18 at 16:07
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Yes, when an irreducible cubic has only one real root, Cardano's method successfully finds it using real square roots and real cube roots.

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  • $\begingroup$ To be precise, one real square root and two real cube roots. $\endgroup$ – Ewan Delanoy May 8 '18 at 16:17

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