Finding $\binom n0+\binom n3+\binom n6+\cdots $ How to get
$$\binom n0 + \binom n3 + \binom n6 + \cdots$$
MY ATTEMPT  
$$(1+\omega)^n = \binom n0 + \binom n1 \omega^1 + \binom n2 \omega^2 + \cdots$$
$$(1+\omega^2)^n = \binom n0 + \binom n1 \omega^2 + \binom n2 \omega^4 + \cdots $$
$$(1 + 1)^n = 2^n = \binom n0 + \binom n1 + \binom n2 + \cdots$$
$$(1+\omega)^n + (1+\omega^2)^n + (1 + 1)^n = 3 \left(\binom n0 + \binom n3 + \binom n6 + \cdots\right)$$
But how to solve LHS? I got the required equation in RHS

Let $\omega = \dfrac{-1+ i\sqrt 3} 2 = \cos120^\circ + i\sin120^\circ$.
Then $\omega^3 = 1$, and $1+\omega = \cos60^\circ + i\sin60^\circ$, so  $(1+\omega)^2 = \omega$.
A bit of arithmetic shows that $n\mapsto(1+\omega)^n + (1+\omega^2)^n$ is a periodic function with period $6$:
\begin{array}{c|c}
n & (1+\omega)^n + (1+\omega^2)^n \\
\hline
0 & \phantom{+}2 \\
1 & \phantom{+}1 \\
2 & -1 \\
3 & -2 \\
4 & -1 \\
5 & \phantom{+}1
\end{array}
Therefore
$$
\binom n 0 + \binom n 3 + \binom n 6 + \cdots = \frac{2^n + \text{a periodic term never exceeding $2$ in absolute value} } 3.
$$
