<s>[INST] 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
 [/INST] $$(1+\omega)^n+(1+\omega^2)^n+2^n\\
=(-\omega^2)^n+(-\omega)^n+2^n\\
=(-1)^n(\omega^{2n}+\omega^n)+2^n\\
$$
i) $n=3m$:
$$(-1)^n(\omega^{2n}+\omega^n)+2^n=2\cdot(-1)^n+2^n$$
ii) $n=3m+1$ or $3m+2$:
$$(-1)^n(\omega^{2n}+\omega^n)+2^n=(-1)^{n+1}+2^n$$
</s>