Compute an approximation of some cubic polynomial over an extension of the $3$-adics Let $K = \mathbb{Q}_3(\sqrt[4]{-3},i)$ and $L = K(\alpha)$
where $$
\alpha = \frac{(1-\zeta_3)(1-\sqrt{-7})}{6}
$$
and $\zeta_3 \in K$ is a primitive 3rd root of unity.
Furthermore, let $v$ be the valuation on $L$ with $v(3)=1$.

Question: Is there a unit $\epsilon \in L^\times$ satisfying the equation $\epsilon^3 \equiv \frac{1}{4}$ modulo an element of valuation $\frac{9}{4}$?

Ideas and Approaches:


*

*I tried to use Hensel's Lemma on the polynomial $f(X)= 4X^3-1$. However, since $f'(X) = 12X^2$ vanishes modulo $3$, it cannot be applied.

*By using Magma, my colleague found out that $\alpha^3 \in K$, i.e. the minimal polynomial of $\alpha$ over $K$ is $x^3-\alpha^3$.

*It is $v(\alpha) = -\frac{1}{2}$, so $\tilde{\alpha} = (1-\zeta_3) \alpha$ is a unit in $L$ since $v(1-\zeta_3)=\frac{1}{2}$. Maybe this can be used for constructing an appropriate $\epsilon$.


Now I ran out of ideas. Could you please help me with this problem? Thanks in advance!
Edit: After skimming through the answers, I noticed that I took the wrong element $\alpha$. In this post, I changed it to the one I intended to use. I will check the answer here more thoroughly though. 
 A: *

*The uniformizer of $L$ is $(-3)^{1/4}$

*If $\epsilon^3 = \frac{1}{4}\bmod (-3)^{9/4}$ for some $\epsilon\in L$ then $\epsilon-1$ is a root of $$(x+1)^3-1/4 =x^3+3 x^2+3x+3-9\bmod (-3)^{9/4}$$

This implies that $v(\epsilon-1)=1/3$ which is a contradiction  since $L/\Bbb{Q}_3$ is tamely ramified with ramification degree $4$, no element of $L$ has valuation $1/3$

Also note that $$v{1/3\choose n}= v(\prod_{k=0}^{n-1} (1/3-k))-v(n!) = -n-\sum_{k\ge 1} \lfloor n/3^k  \rfloor\le -n-n\frac{3^{-1}}{1-3^{-1}}$$ thus $(1+x)^{1/3}=\sum_n {1/3\choose n}x^n$ converges for $v(x)>1+\frac1{3-1}$ and hence $\epsilon^3=1/4\bmod 3^2$ gives that $(4\epsilon^3)^{1/3}\in L, 4^{1/3}\in L$ which is a contradiction since no element of $L$ has valuation $1/3$. The convergence of the binomial series can be stated in term of Hensel lift, when $f'(a)=0\bmod \pi$  we need to use the higher derivatives.
A: First, notice that $\sqrt{-7}$ is already in $\Bbb Q_3(i)$, since $X^2+7\equiv X^2+1\pmod3$: the latter factors in $\Bbb Q_3(i)$, so the former must as well. Thus $L=K$. This field has ramification index $e=4$ over $\Bbb Q_3$, and the residue-field extension degree is $f=2$. For a prime element, we can take $\pi=\sqrt[4]{-3}$.
Second, since you’re talking about units, and $2$ is one such, finding an $\epsilon$ of the type you want is equivalent to finding a unit $\delta$ with such that $v(\delta^3-2)\ge\frac94$: just take $\epsilon=\delta/2$.
We may assume that such a $\delta$ will be $\equiv-1\pmod\pi$: otherwise multiply by a cube root of unity. Thus we write $\delta=-1+\rho$ with $v(\rho)>0$, and expand $\delta^3-2=-1+3\rho-3\rho^2+\rho^3-2$. Let’s look carefully:
$$
\delta^3-2=-3+3\rho-3\rho^2+\rho^3
$$
In case $v(\rho)<\frac13$, the $v$-value of the whole sum is $3v(\rho)<1$, while if $v(\rho)>\frac13$, the $v$-value of the whole is $1$, the minimum valuation being taken on at the first term in the sum. Of course these two are the only possible cases, since there are no elements of $v$-value $1/3$. In other words, it’s not possible for $v(\delta^3-2)$ to take on the value $9/4$, similarly for $v(\epsilon^3-\frac14)$.
