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: Since $a$ is a root of the quadratic, you know $a^2 -3a +1 = 0$, hence, $a^2 = 3a-1$.
This means any power of $a$ can be expressed as a linear combination of $a$ and $1$ in a natural fashion.
E.g. $a^3 = a(3a-1) = 3a^2 -a = 3(3a-1) -a = 8a -3$.
A: $a^2=3a-1$ then
$$\frac{a^3}{a^6 + 1}
\\=\frac{a^3}{(a^2+1)(a^4-a^2 + 1)}
\\=\frac{a^3}{(3a)(9a^2-6a+1-3a+1+1)}
\\=\frac{a^2}{(3)(9a^2-9a+3)}
\\=\frac{a^2}{(9)(3a^2-3a+1)}
\\=\frac{a^2}{(9)(9a-3-3a+1)}
\\=\frac{a^2}{(9)(6a-2)}
\\=\frac{3a-1}{(18)(3a-1)}
\\=\frac{1}{18}
$$
A: 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
$$\frac{a^3}{a^6 + 1} = \frac{a}{a^2 +1}\cdot \frac{a^2}{a^4 - a^2 + 1} = \frac{1}{3}\cdot \frac{1}{6} = \frac{1}{18}.$$
A: As $a\ne0,$ we have $\displaystyle a^2-3a+1=0\iff a^2+1=3a\implies a+\frac1a=3$
and $\displaystyle a^3+\dfrac1{a^3}=\left(a+\frac1a\right)^3-3\left(a+\frac1a\right)^3$
$$\text{Again, }\frac{a^3}{a^6+1}=\frac1{a^3+\dfrac1{a^3}}$$
