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


Since
$$
\frac{1}{3}=\frac{12}{36}=\frac{12a^2}{(a^2+a+1)^2}=\frac{12a^2}{a^4+a^2+1+2a(a^2+a+1)}=\frac{12a^2}{(a^4+a^2+1)+12a^2},
$$
we have
$$
\frac{1}{2}=\frac{1}{3-1}=\frac{12a^2}{[(a^4+a^2+1)+12a^2]-12a^2}=\frac{12a^2}{a^4+a^2+1}.
$$
Hence
$$
\frac{a^2}{a^4+a^2+1}=\frac{1}{24}.
$$
