Let ${(1+x+x^2)}^n=a_0+a_{1}x+a_2{x}^2+.... +a_{2n}{x}^{2n}$ , then   $a_0+a_3+a_6+a_9+\dots$
(until all possible )is equal to
 

$\left(1+\mathrm{x}+{\mathrm{x}}^2\underset{\_}{\underset{\_}{\mathrm{n}}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\dots {\mathrm{a}}_{2\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\right.$
Putting x =1 , $\mathrm{\omega },{\mathrm{\omega }}^2=$ in turn we get, 
$\begin{array}{l}1={\mathrm{a}}_0+{\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+\dots +{\mathrm{a}}_{2\mathrm{n}}\\ 0={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{\omega }+{\mathrm{a}}_2{\mathrm{\omega }}^2+\dots +{\mathrm{a}}_{2\mathrm{n}}{\mathrm{\omega }}^{2\mathrm{n}}\\ \text{ and }0={\mathrm{a}}_0+{\mathrm{a}}_1{\mathrm{\omega }}^2+{\mathrm{a}}_2{\mathrm{\omega }}^4+\dots +{\mathrm{a}}_{2\mathrm{n}}{\mathrm{\omega }}^{2\mathrm{n}}\\ \Rightarrow 3^{\mathrm{n}}=3\left({\mathrm{a}}_0+{\mathrm{a}}_3+{\mathrm{a}}_6+\dots \right)\left(\text{ as }1+\mathrm{\omega }+{\mathrm{\omega }}^2=0,{\mathrm{\omega }}^{3\mathrm{n}}=1\right)\\ {\mathrm{a}}_0+{\mathrm{a}}_3+{\mathrm{a}}_6+\dots =3^{\mathrm{n}-1}\\ \end{array}$