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 $$\frac{a}{a^2+a+1}=\frac{1}{6}$$


The second way is also a bit weird, you can divide by $a$ and $a^2$ and notice that $(a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2} + 2$.
So you have $$\frac{1}{t+1}=\frac{1}{6}$$ and $$\frac{1}{t^2-1}=?$$
