$$\begin{array}{l} \text { 9. Given } (1+4 \sqrt[3]{2}-4 \sqrt[3]{4})^{n+1} \\ =\left(a_{0}+b_{0} \sqrt[3]{2}+c_{0} \sqrt[3]{4}\right)(1+4 \sqrt[3]{2} \\ -4 \sqrt[3]{4}) \end{array}$$

we get
$$a_{n+1}=a_{0}-8 b_{n}+8 c_{n}$$

Since $a_{0}=1$, all $a_{0}$ are odd.
Every non-zero integer $k$ can be expressed as $k=2^{p} \cdot k'$, where $k'$ is odd and $p \geqslant 0$. Define $v(k)=p$. Let $\beta_{0}=v\left(b_{n}\right), \quad \gamma_{0}=v\left(c_{0}\right)$.
(1) For non-negative integer $t, \beta_{2^{t}}=\gamma_{2 t}=t+2$.
Prove: When $t=0$, $b_{1}=4, c_{1}=-4$.

The above assertion holds. Assume $\beta_{2 t}=\gamma_{2 t}=t+2$, then since
$$\begin{aligned} & \left(a+2^{t+2}(b \sqrt[3]{2}+c \sqrt[3]{4})\right)^{2} \\ = & a^{2}+2^{t+3} a(b \sqrt[3]{2}+c \sqrt[3]{4})+2^{2t+4} \\ & (b \sqrt[3]{2}+c \sqrt[3]{4})^{2} \\ = & A+2^{t+3}(B \sqrt[3]{2}+C \sqrt[3]{4}) . \end{aligned}$$

When $a, b, c$ are odd, $A, B, C$ are odd, thus $\beta_{2^{t+1}}=\gamma_{2^{t+1}}=t+3$, hence the assertion holds for all non-negative integers.
(2) If $n, m$ are integers, $b_{n}, c_{n}, b_{m}, c_{m}$ are non-zero, $\beta_{n}=\gamma_{0}=\lambda, \beta_{m}=\gamma_{m}=\mu, \mu<\lambda$, then $b_{n+m}, c_{n+m}$ are non-zero, and $\beta_{n+m}=\gamma_{n+m}=\mu$.
Prove: By $\left(a'+2^{\lambda}\left(b' \sqrt[3]{2}+c' \sqrt[3]{4}\right)\right)$
$$\begin{aligned} \left(a''\right. & \left.+2^{\mu}\left(b'' \sqrt[3]{2}+c'' \sqrt[3]{4}\right)\right) \\ & =A+2^{\mu}(B \sqrt[3]{2}+C \sqrt[3]{4}), \end{aligned}$$

When $a', a''; b', b'', c', c''$ are odd, $A$, $B, C$ are all odd.
(3) For every integer $n \geqslant 1$, let
$$\begin{array}{l} n=2^{t_{r}}+2^{t_{r-1}}+\cdots+2^{t_{1}}, \\ 0 \leqslant t_{1}<\cdots<t_{r}, \end{array}$$

then $c_{n}$ is non-zero, and $\gamma_{n}=t_{1}+2$.
Proof: This follows from (1) and (2).
From (3), the conclusion of the problem is established.
Note: $b_{1080}$ and $c_{1980}$ are divisible by 4 but not by 8.