= Solution
The squared Vandermonde product defining the <polynomial discriminant> is symmetric in all roots. By the fundamental theorem of <symmetric polynomials>, it is a <polynomial> in the elementary symmetric functions, which are the signed coefficients of the monic <polynomial>. Hence it is a <polynomial> function of those coefficients over the base field.
For $f=X^3+pX+q$, $\Delta(f)=-\operatorname{Res}(f,f')$. Eliminating the two leading rows in the Sylvester determinant reduces the resultant to
$$
\operatorname{Res}(f,f')=\det\begin{pmatrix}-2p&-3q&0\\0&-2p&-3q\\3&0&p\end{pmatrix}
=4p^3+27q^2.
$$
Thus
$$
\boxed{\Delta(X^3+pX+q)=-4p^3-27q^2.}
$$
For $X^3-3X+1$, the only possible rational roots are $\pm1$, and neither is a root. The cubic is therefore irreducible, so its <Galois group> is a transitive subgroup of $S_3$. Its discriminant is $81=9^2$. The Vandermonde product changes sign under odd permutations, and its square being a rational square implies that the product itself is rational; hence all Galois permutations are even. The only transitive subgroup of $A_3$ is $A_3$ itself. Consequently, writing $L$ for its splitting field,
$$
\boxed{\operatorname{Gal}(L/\mathbb Q)\cong C_3.}
$$
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