If ${\left(1+x+x^2\right)}^n=a_0+a_{1}x+a_2{x}^2+\dots .+a_{2n}{x}^{2n}$

then the value of  $a_0+a_3+a_6+\dots$ is 

We have, 

 $\begin{array}{r}{\left.1-x+x^2\right)}^n=\left(a_0+a_3{x}^3\right.\left.+a_6{x}^6+\dots \right)\\ \end{array}$ $+x\left(a_1+a_4{x}^3+\dots \right)+x^2\left(a_2+a_5{x}^3+\right)$

Putting $x = 1$  and  $\omega$  and ${\omega }^2$  respectively, we get

 $\begin{array}{l}{3}^n=\left(a_0+a_3+a_6+\dots \right)+\left(a_1+a_4+\dots \right)+\left(a_2+a_5+\dots \right) \left(i\right)\\ 0=\left(a_0+a_3+a_6+\dots \right)+\omega \left(a_1+a_4+\dots \right)+{\omega }^2\left(a_2+a_{5 }+\dots \right) \left(ii\right)\\ 0=\left(a_0+a_3+a_6+\dots \right)+{\omega }^2\left(a_1+a_4+\dots \right)+\omega \left(a_2+a_5+\dots \right) \left(iii\right)\end{array}$

Adding these, we get 

$3^n=3\left(a_0+a_3+a_6+\dots \right)\Rightarrow a_0+a_3+a_6+\dots =3^{n-1}$

Multiplying (ii) and (iii) by ${\omega }^2$ and    $\omega$  respectively and adding

(i), (ii) and (iii), we get 

$a_1+a_4+a_7+\dots =3^{n-1}$

Similarly, we have 

$a_2+a_5+a_8+\dots =3^{n-1}$