Isomorphic extensions of $\mathbb Q_p$ Suppose you are given two cubic Eisenstein polynomials in $\mathbb Q_p[X]$, say $f(X)=X^3-p$ and $g(X)=X^3-pX^2-p^2X-p$. If $\alpha$ is a root of $f(X)$ and $\beta$ is a root of $g(X)$, then the extensions $\mathbb Q_p[\alpha]$ and $\mathbb Q_p[\beta]$ have degree $3$ over $\mathbb Q_p$, and are totally ramified. 

What strategy can one follow to determine those primes $p$ such that $\mathbb Q_p[\alpha]$ and $\mathbb Q_p[\beta]$ are isomorphic? 

If the extensions were unramified the answer is very simple since any two unramified extensions of $\mathbb Q_p$ of the same degree are isomorphic. However, I don't know how to approach the case of totally ramified extensions. I've tried supposing that $\beta\in\mathbb Q_p[\alpha]$ and writing $\beta=a+b\alpha+c\alpha^2$ for some $a,b,c\in\mathbb Q_p$, and then finding relation between $a,b,c$ and the coefficients of the minimal polynomial $g(X)$ of $\beta$, but it led me to nowhere. 
 A: For $p\neq 3$, a totally ramified cubic extension of $\mathbb{Q}_p$ is always given by a cube root.
To see this, let $v$ denote the $p$-adic valuation with $v(p)=1$, $K$ be such an extension, $h(X)$ a monic Eisenstein polynomial generating $K=\mathbb{Q}_p(\beta)$, with $\beta$ a root of $h$. Fix a cube root $p^{1/3}$. Write $h(X) = X^3+pa_2X^2+pa_1X-pa_0$, with $a_i\in \mathbb{Z}_p$ and $v(a_0)=0$, then $$h(p^{1/3}X) = p(X^3+p^{2/3}a_2X+p^{1/3}a_1X-a_0)$$
so $p^{-1/3}\beta$ satisfies polynomial in the bracket, whose reduction is $X^3-\overline{a_0}$. Choose $\alpha \in \overline{\mathbb{Q}_p}$ such that $\alpha^3 = a_0$.
Then for a cube root of unity $\zeta$, $\overline{p^{-1/3}\beta} = \overline{\alpha\zeta}$, so $v(p^{-1/3}\beta - \alpha\zeta)>0$. Any two conjugates over $\mathbb{Q}_p$ of $p^{1/3}\alpha\zeta$ differ by a cube root of unity, since $p\neq 3$, their difference has valuation $1/3$. But
$$v(\beta -p^{1/3}\alpha\zeta) > 1/3$$
Krasner's lemma says $p^{1/3}\alpha\zeta \in \mathbb{Q}_p(\beta)=K$. Note that $(p^{1/3}\alpha\zeta)^3 = pa_0$, so $K\cong \mathbb{Q}_p(\sqrt[3]{pa_0})$.
Above argument shows that a tamely, totally ramified extension of degree $n$ over $\mathbb{Q}_p$ is given by $n$-th root of any Eisenstein polynomial's constant term.

For your question, $K\cong \mathbb{Q}_p(\sqrt[3]{pa_0})$ says immediately your two extensions are isomorphic when $p\neq 3$.
They are not isomorphic when $p=3$. Hint: discriminants of $f(X)$ and $g(X)$ are respectively  $-3^5, -2^2 3^4 5$.
