This is not true in general. For $b=c^3+a^2$ (which is generically not a square), the Galois group shrinks to the order 12 dihedral group $D_6$, implying there is a cubic subfield (other than the one you mention, since of course $a^2-b$ would be a cube); concretely: a root field of $x^3+3cx+2a$.
[EDIT: In view of a late edit of the question, excluding the above family of counterexamples, let me just add that $a=1$, $b=4/5$ is an example not comprised in the above, but also with Galois group isomorphic to $D_6$.]
EDIT: Since the question may have been asked with a different intention, here's a suggestion how to show uniqueness of the intermediate field in the case where it holds. Concretely, assuming irreducibility of the polynomial, existence of a 3-cycle in the Galois group is sufficient (in fact, also necessary), and this can be verified without computing the whole group. E.g., if we're confident that it holds then via Chebotarev's density theorem, one will eventually find a prime $p$ such that the mod-$p$ reduction is separable with exactly $3$ roots in $\mathbb{F}_p$, which does the trick. Alternatively, one can hope to find a ramified prime whose inertia group is generated by a $3$-cycle (this holds ``generically", namely for the prime ideal generated by $b-a^2$ of the function field $F(b)$ with $F=\mathbb{Q}(a)$ (and $a,b$ transcendental); so it will also hold for many, although not for all specializations). For example some discriminant analysis shows that it will hold when $a,b\in \mathbb{Z}$ are chosen coprime and $b-a^2$ has at least one prime divisor $p>3$ of multiplicity coprime to $3$.
