If  $a^i (i = 0, 1, 2, . . . 16)$are real constants such that for every real value of x ${\left(1+x+x^2\right)}^8=a_0+a_{1}x+a_2{x}^2+\dots +a_{16}{x}^{16}$ then $a_{5_{ }}$is  

equal to

${\left(1+x+x^2\right)}^8=\sum _{\genfrac{}{}{0ex}{}{p,q,r\ge 0}{P+q+r=8}} \frac{8!}{p!q!r!}{1}^p{x}^q{\left(x^2\right)}^r$

For , $a_5$we require, $q + 2r = 5$

$\Rightarrow q=5,r=0,p=3$or $q=3,r=1,p=4$

or $q=1,r=2,p=5.$

Thus, coefficient of $x^5$is

$\frac{8!}{5!3!}+\frac{8!}{3!1!4!}+\frac{8!}{1!2!5!}=504$