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.

So
$$a^2 + 1 = 3a$$
and this gives:
$$\frac{a}{a^2 + 1} = \frac{1}{3},$$
and
$$(a^2 + 1)^2 = 9a^2 \implies a^4 + 1 = 7a^2.$$
So
$$\frac{a^2}{a^4 - a^2 + 1} = \frac{a^2}{6a^2} = \frac{1}{6}.$$ And finally
$$\frac{a^3}{a^6 + 1} = \frac{a}{a^2 +1}\cdot \frac{a^2}{a^4 - a^2 + 1} = \frac{1}{3}\cdot \frac{1}{6} = \frac{1}{18}.$$
