Powers of $a$

From what is given:

$\begin{aligned} \frac a{a^2+1} & = \frac 13 \\ \frac {a^2+1}a & = 3 \\ \implies a + \frac 1a & = 3 \\ \left(a + \frac 1a\right)^2 & = 3^2 \\ a^2 + 2 + \frac 1{a^2} & = 9 \\ \implies a^2 + \frac 1{a^2} & = 7 \\ \left(a + \frac 1a\right)^3 & = 3^3 \\ a^3 + 3a + \frac 3a + \frac 1{a^3} & = 27 \\ a^3 + 3(3) + \frac 1{a^3} & = 27 \\ \implies a^3 + \frac 1{a^3} & = 18 \end{aligned}$

Therefore, we have:

$\begin{aligned} x & = \frac {a^3}{a^6+a^5+a^4+a^3+a^2+a+1} & \small \color{#3D99F6} \text{Divide up and down by }a^3 \\ & = \frac 1{a^3+a^2+a+1 + \frac 1a + \frac 1{a^2} +\frac 1{a^3}} \\ & = \frac 1{18+7+3+1} \\ & = \boxed{\dfrac 1{29}} \end{aligned}$