Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$ Is there an easy to solve the problem? The way I did it is to find the value of $a$ from the second expression and then use it to find the value of the first expression. I believe there must be an simple and elegant approach to tackle the problem. Any help is appreciated.

Find the value of $$\frac{a^2}{a^4+a^2+1}$$ if $$\frac{a}{a^2+a+1}=\frac{1}{6}$$


Hint
$$\frac{a^2}{a^4+a^2+1}=\frac{a}{
(a^2+a+1)}\frac{a}{(a^2-a+1)}=\frac{1}{(\frac{a^2+1}{a}+1)}\frac{1}{(\frac{a^2+1}{a}-1)}=\frac{1}{((\frac{a^2+1}{a})^2-1)}$$
