For non-negative integer $n$, write
$$(1+4\sqrt[3]2-4\sqrt[3]4)^n=a_n+b_n\sqrt[3]2+c_n\sqrt[3]4$$
where $a_n,b_n,c_n$ are integers. For any non-negative integer $m$, prove or disprove
$$2^{m+2}|c_n\iff2^m|n$$
So far I have $$\left[\begin{array}{c}a_n\\b_n\\c_n\end{array}\right]=\left[\begin{array}{ccc}1&-8&8\\4&1&-8\\-4&4&1\end{array}\right]\left[\begin{array}{c}a_{n-1}\\b_{n-1}\\c_{n-1}\end{array}\right].$$
On
$$ x \equiv \sqrt[3]{2} \quad\Longrightarrow\quad x^{2} = \sqrt[3]{4},\ x^{3} = 2,\ x^{4} = 2x,\ x^{5} = 2x^{2}, x^{6} = 4,\ x^{7} = 4x\ldots $$
$$\left(1 + 4\sqrt[3]{2} - 4\sqrt[3]{4}\right)^{n} = \left(1 + 4x - 4x^{2}\right)^{n} = a_{n} + b_{n}x + c_{n}x^{2} $$
$$ {1 \over 1- z\left(1 + 4x - 4x^{2}\right)} = \overbrace{\sum_{n = 0}^{\infty}z^{n}a_{n}}^{\equiv\ {\rm A}\left(z\right)}\ +\ x\quad\overbrace{\sum_{n = 0}^{\infty}z^{n}b_{n}}^{\equiv\ {\rm B}\left(z\right)}\ +\ x^{2}\quad\overbrace{\quad\sum_{n = 0}^{\infty}z^{n}c_{n}}^{\equiv\ {\rm C}\left(z\right)} $$
$$ 1= \left\lbrack\left(1- z\right) - 4zx + 4zx^{2}\right\rbrack \left\lbrack {\rm A}\left(z\right) + {\rm B}\left(z\right)x + {\rm C}\left(z\right)x^{2} \right\rbrack $$
$$ \left\lbrace% \begin{array}{rcrcrcl} \left(1 - z\right){\rm A}\left(z\right) & + & 8z{\rm B}\left(z\right) & - & 8z{\rm C}\left(z\right) & = & 1 \\[1mm] -4z{\rm A}\left(z\right) & + & \left(1 - z\right){\rm B}\left(z\right) & + & 8z{\rm C}\left(z\right) & = & 0 \\[1mm] 4z{\rm A}\left(z\right) & - & 4z{\rm B}\left(z\right) & + & \left(1 - z\right){\rm C}\left(z\right) & = & 0 \end{array}\right. $$
$$ \left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle = \left\vert\Psi_{0}\right\rangle $$
$$ {\bf M} \equiv\left(% \begin{array}{rrr} 1 & -8 & 8 \\ 4 & 1 & -8 \\ -4 & 4 & 1 \end{array}\right)\,, \qquad \left\vert\Psi\left(z\right)\right\rangle \equiv \left(% \begin{array}{c} {\rm A}\left(z\right) \\[1mm] {\rm B}\left(z\right) \\[1mm] {\rm C}\left(z\right) \end{array}\right) $$
$$ \left\vert\Psi_{0}\right\rangle \equiv \left(% \begin{array}{c} 1 \\[1mm] 0 \\[1mm] 0 \end{array}\right)\,, \qquad \left\vert\Psi_{1}\right\rangle \equiv \left(% \begin{array}{c} 0 \\[1mm] 1 \\[1mm] 0 \end{array}\right)\,, \qquad \left\vert\Psi_{2}\right\rangle \equiv \left(% \begin{array}{c} 0 \\[1mm] 0 \\[1mm] 1 \end{array}\right)\,, $$ and $$ \left(% \begin{array}{c} a_{n} \\ b_{n} \\ c_{n} \end{array}\right) = \left.% {1 \over n!}\,{{\rm d}^{n}\left\vert\Psi\left(z\right)\right\rangle \over {\rm d}z^{n}} \right\vert_{z = 0} $$
Since
\begin{align} \left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle = \left\vert\Psi_{0}\right\rangle &\quad\Longrightarrow& \left\vert\Psi\left(0\right)\right\rangle = \left\vert\Psi_{0}\right\rangle \\[1mm] -{\bf M}\left\vert\Psi\left(z\right)\right\rangle + \left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle' = 0 &\quad\Longrightarrow& \left\vert\Psi\left(0\right)\right\rangle' = {\bf M}\left\vert\Psi\left(0\right)\right\rangle = {\bf M}\left\vert\Psi_{0}\right\rangle \\[1mm] -2{\bf M}\left\vert\Psi\left(z\right)\right\rangle' + \left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle'' = 0 &\quad\Longrightarrow& \left\vert\Psi\left(0\right)\right\rangle'' = 2{\bf M}\left\vert\Psi\left(0\right)\right\rangle' = 2{\bf M}^{2}\left\vert\Psi_{0}\right\rangle \\[1mm] -3{\bf M}\left\vert\Psi\left(z\right)\right\rangle'' + \left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle''' = 0 &\quad\Longrightarrow& \left\vert\Psi\left(0\right)\right\rangle''' = 3{\bf M}\left\vert\Psi\left(0\right)\right\rangle' = 3!\,{\bf M}^{3}\left\vert\Psi_{0}\right\rangle \\ \vdots&\vdots&\vdots \end{align} we'll get $$ \left(% \begin{array}{ccc} a_{n} \\ b_{n} \\ c_{n} \end{array}\right) = {1 \over n!}\,n!\,{\bf M}^{n}\left\vert\Psi_{0}\right\rangle \quad\Longrightarrow\quad c_{n} = \left\langle\Psi_{2}\left\vert{\bf M}^{n}\right\vert\Psi_{0}\right\rangle $$
In addition, ${\bf M} = 1 + 4{\bf Q}$ where $$ {\bf Q} \equiv\left(% \begin{array}{rrr} 0 & -2 & 2 \\ 1 & 0 & -2 \\ -1 & 1 & 0 \end{array}\right) \quad\mbox{such that}\quad {\bf M}^{n} = \sum_{\ell = 0}^{n}{n \choose \ell}2^{2\ell}{\bf Q}^{\ell} $$
On
Seems to me that this is a $2$-adic question. Our given number, let’s call it $\rho$ for convenience, is of the form $1+4z$, and we’re looking at its multiplicative properties. But, calling $R=\mathbb Z_2(\root 3\of 2\,)$, the multiplicative group $1+4R$ is isomorphic (via the $2$-adic logarithm) to an open subgroup of the additive group $R^+$. Thus $1+4R$ is a free $\mathbb Z_2$-module of rank three. Clearly $1+4$, $1+4\root3\of2$, and $1-4\root3\of2^2$ form a basis of our multiplicative module, just look at their logarithms. Notice also that $(1+4\root3\of2)(1-4\root3\of2^2\,)=\rho-32$, and this means that $\rho^n$ is very close to $(1+4\root3\of2\,)^n(1-4\root3\of2^2)^n$, $2$-adically speaking. But we know the exact $2$-adic valuation of $ (1+4\root3\of2\,)^n-1$ and of $ (1-4\root3\of2^2)^n-1$, for any $n\ge1$: $$ v\left[(1+4\root3\of2\,)^n-1\right]=v(n)+2+\frac13\>,\> v\left[(1-4\root3\of2^2)^n-1\right]=v(n)+2+\frac23\,. $$ The desired result follows, after a few details are supplied.
EDIT:
Rather than replace the unnecessarily complicated and (probably) confusing explanation above, let me give a clearer argument. The desired result follows from the fact that for $z\in R$, we have the congruence $(1+4z)^n\equiv1+4nz\pmod{8n}$, using $z=\root3\of2-\root3\of2^2$. Although this can be proved directly from the Binomial Theorem, it’s quicker and easier to see it by using the explicit isomorphism between the multiplicative group $1+4R$ and the additive group $4R$ given by the logarithm and exponential. We have: $$ \log(1+4z)=4\left[z-2z^2+\frac{16z^3}3+\cdots\right] $$ and $$ \exp(4z)=1+4\left[z+2z^2+\frac{8z^3}3+\cdots\right]\,, $$ where the two series shown in brackets are inverses of each other, and the logarithmic one clearly has all coefficients $2$-adic integers, indeed all beyond the first are divisible by $2$. As a result, the same is true for the exponential series. So you see that in the original notation, $b_n\equiv 4n\pmod{8n}$ and $c_n\equiv-4n\pmod{8n}$.
Collecting my thoughts. This is in dire need of some streamlining and condensing, but I don't have enough time now.
We are working inside the ring $O$ of algebraic integers of the field $K=\mathbb{Q}(\root3\of2)$ that has a monomial integral basis $\{1,\root3\of2,\root3\of4\}$. The trace function of $K$ is easily seen to be $$ tr(A+B\root3\of2+C\root3\of4)=3A $$ for all rationals $A,B,C$. A consequence of particular interest to us is the formula $$ C=tr\left(\frac1{3\root3\of4}(A+B\root3\of2+C\root3\of4)\right) $$ that allows us to extract the $C$-coefficient.
The number $\varpi=\root3\of2$ generates a prime ideal $P$ of $O$. We are concerned about the effect of divisibility by a power of $\varpi$ on the trace. Let us denote by $\nu_{\varpi}(z)$ the exponent of the highest power of $\varpi$ that divides the number $z\in K$. Let us similarly denote by $\nu_2(A)$ the exponent of the highest power of two dividing the integer $A$. As $2=\varpi^3$ we see that for a number $z=A+B\root3\of2+C\root3\of4\in O$ we have $$ \nu_\varpi(z)=\min\{3\nu_2(A),3\nu_2(B)+1,3\nu_2(C)+2\}. $$ As a Corollary of this we see that if for some integer $k$ we have $\nu_\varpi(z)>3k$, then $\nu_2(tr(z))>k$. Another important observation is that if $\nu_\varpi(z_1)<\nu_\varpi(z_2)$, then $\nu_\varpi(z_1+z_2)=\nu_\varpi(z_1)$.
The question is about the powers of the number $x=1+4\root3\of2-4\root3\of4=1+\alpha$, where $$ \alpha=4\root3\of2-4\root3\of4=\varpi^7(1-\varpi). $$ By the above observations we have $$ c_n=tr(\frac1{3\root3\of4}x^n). $$ A key observation is the following. Expanding $x^n$ by the binomial formula gives us $$ x^n-1=(1+\alpha)^n-1=n\alpha+{n\choose2}\alpha^2+\cdots+\alpha^n. $$ As $\nu_{\varpi}(\alpha)=7$ and $\nu_{\varpi}(2)=3$, it is easy to see that on the r.h.s. the first term dominates in the sense that $\nu_{\varpi}(n\alpha)<\nu_{\varpi}({n\choose k}\alpha^k)$ for all $k>1$. Consequently $$\nu_\varpi(x^n-1)=\nu_\varpi(n\alpha).$$
The reason for isolating that term $1$ from $x$ is that $$ c_n=tr(\frac1{3\root3\of4}x^n)=tr(\frac1{3\root3\of4}(x^n-1))+tr(\frac1{3\root3\of4})=tr(\frac1{3\root3\of4}(x^n-1)), $$ as $tr(1/\root3\of4)=0$. We next see that $$\nu_{\varpi}(\frac1{3\root3\of4}(x^n-1))=\nu_{\varpi}(\frac{n\alpha}{\root3\of4})=7+\nu_{\varpi}(n)-2=5+3\nu_2(n)>3(\nu_2(n)+1).$$ By our earlier observations it follows that $\nu_2(c_n)>\nu_2(n)+1$, and hence that $\nu_2(c_n)\ge\nu_2(n)+2$.
To show that $\nu_2(c_n)$ is exactly $\nu_2(n)+2$, we make the observations that $$ \nu_2(tr(\frac{n\alpha}{\varpi^2}))= \nu_2(n\,tr(\frac{\varpi^7-\varpi^8}{\varpi^2}))= \nu_2(n)+\nu_2(tr(\varpi^5-\varpi^6))=\nu_2(n)+2 $$ as $tr(\varpi^5)=0$ and $tr(\varpi^6)=tr(4)=12=2^2\cdot3$.