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 $$\frac{a^4+a^2+1}{a^2}=24.$$
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
$$ \frac{a^2}{a^4+a^2+1}=\frac{5a-1}{120a-24}=\frac{1 \times\color{red}{(5a-1)}}{24\times\color{red}{(5a-1)}}=\frac1{24}. $$
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
$$B=A^2-2A=6^2-2\cdot6=24.$$
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 $$\frac{1}{t^2-1}=?$$
$$\frac{a^2}{a^4+a^2+1}=\frac{a}{ (a^2+a+1)}\frac{a}{(a^2-a+1)}=\frac{1}{(\frac{a^2+1}{a}+1)}\frac{1}{(\frac{a^2+1}{a}-1)}=\frac{1}{((\frac{a^2+1}{a})^2-1)}$$
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 $$ \frac{a^2}{a^4+a^2+1}=\frac{1}{24}. $$
