Let $P \in K[X]$ be an irreducible cubic polynomial where $K \subset \mathbb C$ and $\alpha$ a root of $P$ and one suppose $K[\alpha]$ is a radical extension of $K$.
Because $[K[\alpha]:K]=3$ ($3$ is prime, so no intermediate field) there exist $\beta \in K[\alpha]-K$ such as $\beta^t \in K$ (with $t$ prime) and $K[\beta]=K[\alpha]$.
Of course $t\geq 3$, but, why $t=3$?
Often, some say it's obvious , because $[K[\beta]:K]=t$, but why degree of $\beta$ would be $t$ ($i^6=1$ and degree of $i$ is $2$, not $6$).
Let $D$ the discriminant of $P$. I If $D$ is a square in $K$, $K[\alpha]$ is not a splitting field of $P$ and $K[\alpha]$ is not Galois over $K$ and $t=3$ ( lemma 2.3b of Isaacs and Moulton, Real fields and Repeated radical extensions, journal of algebra 201, 1998).
But if $D$ is not a square in $K$?
In this case, $K[\alpha]$ is Galois over K, so I can't apply this lemma 2.3b and the 2.3a say only "$K[\alpha]$ contains some roots of unity different from $\pm 1$".
However, if $D$ is a square and $j$ ($j^3=1,j \neq 1$), then $-3D$ is a square in $K$ and we have one $\beta$ with $t=3$.
But, if $D$ is a square, $j$ not in $K$ ??
Note : if $D$ is a square and $[K:\mathbb Q]$ is odd, then $K[\beta]$ is not a radical extension of $K$. Thanks.