= Solution
For a monic <polynomial> with roots $r_1,\ldots,r_n$, its discriminant is
$$
\operatorname{Disc}(f)=\prod_{i<j}(r_i-r_j)^2.
$$
Here $\operatorname{Disc}(g)=(u^3-v^3)^2$. Both $u^3-v^3$ and the Vandermonde product in the $\alpha_i$ are alternating homogeneous cubics. Evaluation at $(\alpha_1,\alpha_2,\alpha_3)=(0,1,-1)$ fixes the constant and gives
$$
(u^3-v^3)^2=-27\prod_{i<j}(\alpha_i-\alpha_j)^2=-27\operatorname{Disc}(f).
$$
Using $\alpha_1+\alpha_2+\alpha_3=0$, $\sum_{i<j}\alpha_i\alpha_j=a$, and $\alpha_1\alpha_2\alpha_3=-b$, direct expansion gives
$$
uv=-3a,\qquad u^3+v^3=-27b.
$$
Consequently
$$
g(X)=X^2+27bX-27a^3,
$$
and
$$
\operatorname{Disc}(g)=729b^2+108a^3=-27(-4a^3-27b^2).
$$
Thus
$$
\boxed{\operatorname{Disc}(f)=-4a^3-27b^2}.
$$
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