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 ?
Yes, when an irreducible cubic has only one real root, Cardano's method successfully finds it using real square roots and real cube roots.