The integer $c_n$ in $(1+4\sqrt[3]2-4\sqrt[3]4)^n=a_n+b_n\sqrt[3]2+c_n\sqrt[3]4$ 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
 $$\left[\begin{array}{c}a_n\\b_n\\c_n\end{array}\right]=\left[\begin{array}{ccc}1&-8&8\\4&1&-8\\-4&4&1\end{array}\right]\left[\begin{array}{c}a_{n-1}\\b_{n-1}\\c_{n-1}\end{array}\right].$$

$$
x \equiv \sqrt[3]{2}
\quad\Longrightarrow\quad
x^{2} = \sqrt[3]{4},\ x^{3} = 2,\ x^{4} = 2x,\ x^{5} = 2x^{2}, x^{6} = 4,\ x^{7} = 4x\ldots 
$$
$$\left(1 + 4\sqrt[3]{2} - 4\sqrt[3]{4}\right)^{n}
=
\left(1 + 4x - 4x^{2}\right)^{n}
=
a_{n} + b_{n}x + c_{n}x^{2}
$$
$$
{1
 \over
 1- z\left(1 + 4x - 4x^{2}\right)}
=
\overbrace{\sum_{n = 0}^{\infty}z^{n}a_{n}}^{\equiv\ {\rm A}\left(z\right)}\
+\
x\quad\overbrace{\sum_{n = 0}^{\infty}z^{n}b_{n}}^{\equiv\ {\rm B}\left(z\right)}\
+\
x^{2}\quad\overbrace{\quad\sum_{n = 0}^{\infty}z^{n}c_{n}}^{\equiv\ {\rm C}\left(z\right)}
$$
$$
1=
\left\lbrack\left(1- z\right) - 4zx + 4zx^{2}\right\rbrack
\left\lbrack
{\rm A}\left(z\right) + {\rm B}\left(z\right)x + {\rm C}\left(z\right)x^{2}
\right\rbrack
$$
$$
\left\lbrace%
\begin{array}{rcrcrcl}
\left(1 - z\right){\rm A}\left(z\right)
& + &
8z{\rm B}\left(z\right)
& - &
8z{\rm C}\left(z\right)
& = &
1
\\[1mm]
-4z{\rm A}\left(z\right)
& + &
\left(1 - z\right){\rm B}\left(z\right)
& + &
8z{\rm C}\left(z\right)
& = &
0
\\[1mm]
4z{\rm A}\left(z\right)
& - &
4z{\rm B}\left(z\right)
& + &
\left(1 - z\right){\rm C}\left(z\right)
& = &
0
\end{array}\right.
$$
$$
\left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle
=
\left\vert\Psi_{0}\right\rangle
$$
$$
{\bf M}
\equiv\left(%
\begin{array}{rrr}
1 & -8 & 8
\\
4 & 1 & -8
\\
-4 & 4 & 1
\end{array}\right)\,,
\qquad
\left\vert\Psi\left(z\right)\right\rangle
\equiv
\left(%
\begin{array}{c}
{\rm A}\left(z\right)
\\[1mm]
{\rm B}\left(z\right)
\\[1mm]
{\rm C}\left(z\right)
\end{array}\right)
$$
$$
\left\vert\Psi_{0}\right\rangle
\equiv
\left(%
\begin{array}{c}
1
\\[1mm]
0
\\[1mm]
0
\end{array}\right)\,,
\qquad
\left\vert\Psi_{1}\right\rangle
\equiv
\left(%
\begin{array}{c}
0
\\[1mm]
1
\\[1mm]
0
\end{array}\right)\,,
\qquad
\left\vert\Psi_{2}\right\rangle
\equiv
\left(%
\begin{array}{c}
0
\\[1mm]
0
\\[1mm]
1
\end{array}\right)\,,
$$
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
$$
{\bf Q}
\equiv\left(%
\begin{array}{rrr}
0 & -2 & 2
\\
1 & 0 & -2
\\
-1 & 1 & 0
\end{array}\right)
\quad\mbox{such that}\quad
{\bf M}^{n}
=
\sum_{\ell = 0}^{n}{n \choose \ell}2^{2\ell}{\bf Q}^{\ell}
$$
