Partial answer: Let $f(x)=x^3+ax+b$, let $K$ be its splitting field, and $\alpha$, $\beta$ and $\gamma$ be the roots of $f$ in $K$.
First of all, $f$ has to be irreducible, which is the same as saying it doesn't have a rational root: if it's not, and say $\alpha$ is rational, then $f(x)$ factors as $(x-\alpha)g(x)$ with $g$ a quadratic polynomial; then $K$ is also the splitting field of $g$, so it has degree $\leq2$.
So let's assume that $f$ is irreducible. Its splitting field $\mathbf Q(\alpha)$ has degree $\deg(f)=3$, so if we want $[K:\mathbf Q]=3$, we need $K=\mathbf Q(\alpha)$. In other words, we want the two other roots $\beta,\gamma$ to be in $\mathbf Q(\alpha)$. Let's look at the relation between roots and coefficients for $f$:
From the first and the third equation, you see that $\beta+\gamma=-\alpha$ and $\beta\gamma=-b/\alpha=\alpha^2+a$, so $\beta$ and $\gamma$ are the roots of the second degree polynomial $g(y)=y^2+\alpha y +\alpha^2+a\in\mathbf Q(\alpha)$. Those roots are in $\mathbf Q(\alpha)$ if, and only if, the discriminant $\Delta=-3\alpha^2-4a$ of $g$ is a square in $\mathbf Q(\alpha)$.
Now, the problem is that I'm not sure how to determine whether $\Delta$ is a square or not...