Let $A_r={\left(x+\frac{1}{x}\right)}^3{\left(x^2+\frac{1}{x^2}\right)}^3{\left(x^3+\frac{1}{x^3}\right)}^3\mathrm{....}{\left(x^r+\frac{1}{x^r}\right)}^3$. If $x^2+x+1=0$ then $\frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+---\infty$

$x^2+x+1=0$
roots are $\omega and {\omega }^2\text{i.e.,}x=\omega \text{(or)}\omega ,{\omega }^2$
$\begin{array}{rl} & {(x+\frac{1}{x})}^3={(\omega +\frac{1}{\omega })}^3=-1\\ & {(x^2+\frac{1}{x^2})}^3={({\omega }^2+\frac{1}{{\omega }^2})}^3=-1\\ & {(x^3+\frac{1}{x^3})}^3={({\omega }^3+\frac{1}{{\omega }^3})}^3=2\end{array}$
$\begin{array}{rl} & \text{Now}{A}_3=(-1{)}^3(-1{)}^3(2{)}^3=8,A_6=8^2,A_9=8^3\\ & \frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\dots \mathrm{\infty }=\frac{1}{8}+\frac{1}{8^2}+\frac{1}{8^3}+\dots \mathrm{\infty }\end{array}$
$\begin{array}{rl} & =\frac{\frac{1}{8}}{1-\frac{1}{8}}\\ & =\frac{1}{7}\end{array}$