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. From the equation, one easily gets 
$$
a^2=5a-1, \quad a^4=(5a-1)^2=25a^2-10a+1=115a-24
$$ giving in the first expression

$$
\frac{a^2}{a^4+a^2+1}=\frac{5a-1}{120a-24}=\frac{1 \times\color{red}{(5a-1)}}{24\times\color{red}{(5a-1)}}=\frac1{24}.
$$ 

