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The polynomial is Eisenstein at every prime number $p$, so a polynomial root $\alpha$ generates a totally ramified cubic field $F/{\mathbb{Q}}_p$. The splitting-field Galois group is a transitive subgroup of $S_3$, hence $C_3$ or $S_3$. Its polynomial discriminant isThe product of the three polynomial root differences is a square root of $\Delta$, transformed by the sign of the polynomial root permutation. Thus the group is $C_3$ exactly when $\Delta$ is a square, and otherwise it is $S_3$. In the latter case the splitting field isIndeed adjoining one polynomial root leaves a quadratic factor, whose polynomial discriminant is $\Delta /f^{\prime }(\alpha {)}^2$, so the square root of $\Delta$ supplies its remaining polynomial roots.

Suppose first that $p>3$. The unit $-(4p+27)$ reduces to $-27$, and Hensel's lemma makes it a square exactly when $-3$ is a square modulo $p$. Quadratic reciprocity givesIf $p\equiv 1(\mathrm{mod}3)$, the field generated by $\alpha$ is already the cyclic splitting field, totally ramified of degree three. If $p\equiv 2(\mathrm{mod}3)$, the nonsquare unit gives an unramified quadratic field $E={\mathbb{Q}}_p(\sqrt{\Delta })$. Since $F$ and $E$ have coprime degrees, $[FE:{\mathbb{Q}}_p]=6$. Its residue degree is at least two and its ramification index at least three; their product is six, so they are exactly two and three. Thus its inertia group is the normal $C_3$ inside $S_3$.

In both cases the ramification index is three and is not divisible by $p$. The wild inertia group is trivial, so all positive lower groups are trivial. As a different check, $f^{\prime }(\alpha )=3{\alpha }^2+p$ has valuation two in $F$ when $p>3$, agreeing with tame different exponent $3-1$.

For $p=3$ the polynomial discriminant is $-351=-3^3\cdot 13$, which is nonsquare. Hence $G=S_3$. The quadratic subfield can be written $E={\mathbb{Q}}_3(\beta )$ with ${\beta }^2=-39$. It is ramified of degree two, by the Eisenstein polynomial $T^2+39$. Together with the totally ramified cubic field this forces $e(L/{\mathbb{Q}}_3)$ to be divisible by both two and three. Since $[L:{\mathbb{Q}}_3]=6$, the splitting field is totally ramified. In its normalized valuation,

All differences between distinct polynomial roots have the same valuation, since $S_3$ permutes the polynomial root pairs and preserves the absolute value on a field. Alsohas valuation six, so each polynomial root difference has valuation three. A nonidentity three-cycle fixes $\beta$, and thereforeThe transposition fixing $\alpha$ negates $\beta$, so sends $\pi$ to $-\pi$ and has displacement valuation $v_L(2\pi )=1$. The other transpositions have the same value by conjugacy. The uniformizer criterion now determines every group:For the wild case the different exponent is $5+2=7$. Independently, the cubic subfield has different exponent $v_F(3({\alpha }^2+1))=3$, and the quadratic extension $L/F$ is tame with exponent one. The tower formula gives $1+2\cdot 3=7$, confirming the filtration. Its positive upper break is $1/2$, because $[G_0:G_1]=2$. This completes ramification of the splitting field of X3 plus pX plus p for every odd prime number.