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

You have
$$(1+\omega)^n+(1+\omega^2)^n+2^n=3\left(\binom n0+\binom n3+\binom n6+\cdots\right)$$
Now note that
$$(1+\omega)^n+(1+\omega^2)^n=(-\omega^2)^n+(-\omega)^n=(-1)^n(\omega^n+\omega^{2n})$$
This is equal to $(-1)^n\cdot 2$ if $n\equiv 0\pmod 3$ or $(-1)^n\cdot (-1)=(-1)^{n+1}$ if $n\not\equiv 0\pmod 3.$
