# Is the unique real root of $X^3-(X+1)$ solvable by real radicals?

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 ?

• 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. – Peter May 8 '18 at 16:07

## 1 Answer

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

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