<s>[INST] 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 $$\frac{a}{a^2+a+1}=\frac{1}{6}$$

 [/INST] You are asked to express $\dfrac1B=\dfrac1{a^2+1+a^{-2}}$ in terms of $\dfrac1A=\dfrac1{a+1+a^{-1}}$.
Squaring "to see", 
$$A^2=(a+1+a^{-1})^2=a^2+1+a^{-2}+2a+2+2a^{-1}=B+2A.$$
This gives us
$$B=A^2-2A=6^2-2\cdot6=24.$$
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