See all problems here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=528313

If $\omega=\frac{-1+i\sqrt{3}}{2}$, then $c_n=\sqrt[3]{2}(\sqrt[3]{2}-1)^n+\omega\sqrt[3]{2}(\omega\sqrt[3]{2}-1)^n+\omega^2\sqrt[3]{2}(\omega^2\sqrt[3]{2}-1)^n$ just because of the symmetry between $\sqrt[3]{2}, \omega\sqrt[3]{2}, \omega^2\sqrt[3]{2}$. So $c_n$ is the solution to some linear recurrence relation with characteristic polynomial $(x-(\sqrt[3]{2}-1))(x-(\omega\sqrt[3]{2}-1))(x-(\omega^2\sqrt[3]{2}-1))=((x+1)^3-2)=x^3+3x^2+3x-1$. So then $c_n=-3c_{n-1}-3c_{n-2}+c_{n-3}$. Clearly this is a periodic sequence $\mod 3$ of order $3$, with $c_0=c_1=0$ and $c_2=1$. So $c_i\equiv1\mod3$ iff $i\equiv2\mod3$, and otherwise $c_i\equiv0\mod3$. P.S. The link is http://www.artofproblemsolving.com/Forum/viewtopic.php?t=528313

Write \begin{align*} a_{n + 3} + b_{n + 3}\sqrt[3]{2} + c_{n + 3}\sqrt[3]{4} &= (a_{n} + b_{n}\sqrt[3]{2} + c_{n}\sqrt[3]{4})[(\sqrt[3]{2} - 1)^3] \\ &= (a_{n} + b_{n}\sqrt[3]{2} + c_{n}\sqrt[3]{4})(7 + 3\sqrt[3]{2} - 3\sqrt[3]{4}) \end{align*} so it's clear that $c_{n + 3} \equiv 7c_n \equiv c_n \pmod{3}$. It remains only to check $c_2, c_3, c_4$ and see that only $c_2 \equiv 1 \pmod{3}$.

$$c_n=3c_{n+1}+3c_{n+2}+c_{n+3}$$$$c_n \equiv2 c_{n+3}\mod3$$

$(\sqrt[3]{2}-1)^3=1-3\sqrt[3]{2}+3\sqrt[3]{4} \implies (\sqrt[3]{2}-1)^{n+3}=(a_n+b_n{\sqrt [3]2}+c_n {\sqrt [3]4})(1-3\sqrt[3]{2}+3\sqrt[3]{4})=a_{n+3}+b_{n+3}{\sqrt [3]2}+(3a_n-3b_n+c_n) {\sqrt [3]4}$