Find $\frac{a^3}{a^6 + 1}$ given $a$ is a root of a quadratic equation 
If $a$ is a root of the equation $x^2 - 3x + 1 = 0$, then find the value of $\frac{a^3}{a^6 + 1}$.

So, I figured we can use the quadratic formula, and formed the following equation:
$$a=\frac{-(-3)+\sqrt{9-4}}{2(1)}\implies a=\frac{3+\sqrt5}2$$
But what I am thinking is, if I begin to find the required value, it will take me hours. And I believe, there must be some shortcut to this question. I had tried to solve this question with the manual process but it took me a lot of squares (one was $2012^2$!) 
Can someone please help me.
Thank you.

$a^2=3a-1$ then
$$\frac{a^3}{a^6 + 1}
\\=\frac{a^3}{(a^2+1)(a^4-a^2 + 1)}
\\=\frac{a^3}{(3a)(9a^2-6a+1-3a+1+1)}
\\=\frac{a^2}{(3)(9a^2-9a+3)}
\\=\frac{a^2}{(9)(3a^2-3a+1)}
\\=\frac{a^2}{(9)(9a-3-3a+1)}
\\=\frac{a^2}{(9)(6a-2)}
\\=\frac{3a-1}{(18)(3a-1)}
\\=\frac{1}{18}
$$
