<s>[INST] Binomial expansion of complex variable For the binomial expansion $ (1+x)^n = p_0 + p_1x+p_2x^2+...,$ where 
$p_i$ refer to the binomial coefficients. 
Need to substitute the cube roots of unity (1, w, $w^2$) for the variable x; with w = $\dfrac{-1+i\sqrt{3}}{ 2},$
find the sums:\begin{align}  (a) p_0 + p_3 + p_6 + ..., \\  (b) p_1 + p_4 + p_7 + ..., \\  (c) p_2 + p_5 + p_8 + ..., \\
\end{align}
I have substituted the values as follows:
$(d)\;2^n     = $$p_0+p_1+p_2+ p_3 + p_4 + p_5+...$
$(e)\;(1+w)^n = $$p_0+p_1.w+p_2.w^2+p_3+p_4.w+p_5.w^2+...$
$(f)\;(1+$$w^2$$)^n =(-w)^n = $$p_0+p_1.w^2+p_2.w+p_3+p_4.^2+p_5.w^2+...$
On adding (d), (e), (f); get the non-multiple of 3 terms vanished by the sum property of the cube roots of unity (1+w+$w^2$=0). Rest (non-zero) terms have each $p_i$ terms' coefficient being 3 (1+1+1=3) for all such terms.
(i)  $(2^n + (1+w)^n + (-w)^n)/3 = p_0 + p_3 + p_6 + ...$ 
Geometrically/algebraically plotting the points (1+w), (1+$w^2$); we get:
1+w=$\dfrac{1+i\sqrt{3}}{2},$ 1+$w^2$=-w=$\dfrac{1-i\sqrt{3}}{2}.$ 
Obtaining the trigonometrical, & then exponential forms for the same, get:
(ii)  1+w=$\dfrac{1+i\sqrt{3}}{2}$= $cis(\dfrac{n\pi}{3})$= $e^\dfrac{in\pi}{3}$
(iii) -w =$\dfrac{1-i\sqrt{3}}{2}$= $cis(\dfrac{-n\pi}{3})$= $e^\dfrac{-in\pi}{3}$
Adding $e^\dfrac{in\pi}{3}$ + $e^\dfrac{-in\pi}{3}$ we get:
2.$\dfrac{e^\dfrac{in\pi}{3} + e^\dfrac{-in\pi}{3}}{2} = 2.\dfrac{cos(n\pi)}{3}$
Hence, get (i) modified as:
(iv)  $p_0 + p_3 + p_6 + ... = \dfrac{2^n + 2cos(\dfrac{n\pi}{3})}{3}$
This solves only the part (a) of the question. 
For the parts (b) & (c), there is no such closed form expression possible for me. 
 [/INST] Ill do part 2. Since all coefficients of $x^{1+3k}$ are to be retained, we multiply by $x^2$ to make the power $x^{3+3k}$, ie a multiple of $3$. 
$$(1+x)^n = a_o + a_1 x + a_2 x^2...a_nx^n$$
$$x^2(1+x)^n = a_ox^2+a_1x^3+a_2 x^4 ... a_nx^{n+2}$$
Now substituting cube roots of unity in above equation leaves required terms intact ($z^3 = 1$, where $z$ is cube root of unity).
All the other terms vanish, ( $1+z+z^2 =0$ if $z$ is cube root of unity)
Let $P(x) = x^2 (1+x)^n$. Then on we have:
$$\begin{align} a_1+a_4+a_7 ...  &= \dfrac{P(1) + P(\omega) + P(\omega^2)}{3} \\
 &= \dfrac{2^n + \omega^2(1+\omega)^n + \omega(1+\omega^2)^n}{3}
\end{align}$$
</s>