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


From the first equation (inverted),
$$\frac{a^2+a+1}a=6$$ or $$\frac{a^2+1}a=5.$$
Then squaring,
$$\frac{a^4+2a^2+1}{a^2}=25$$
or
$$\frac{a^4+a^2+1}{a^2}=24.$$
