What is the value of $\frac{a^3}{a^6+a^5+a^4+a^3+a^2+a+1}$ If $\frac{a}{a^{2}+1} = \frac{1}{3}$ Then find the value of $$\frac{a^3}{a^6+a^5+a^4+a^3+a^2+a+1}$$
Any hints to help me?

Using $a^2=3a-1$ we can compute $a^3=8a-3,\; a^4=21a-8, ... , a^6=144a-55$ by iteration. Then your term equals 
$$
\frac{8a-3}{29(8a-3)}=\frac{1}{29}.
$$
