Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ . 
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .

What I Tried: We have :-
$$ (a^3 - 503a - 500)^5 = [a(a^2 - 3) - 500(a - 1)]^5$$
$$= \Bigg(\Bigg[\frac{1 + \sqrt{2009}}{2}\Bigg]\Bigg[\frac{1009 + \sqrt{2009}}{2}\Bigg] - 500\Bigg[\frac{\sqrt{2009} - 1}{2}\Bigg]\Bigg)^5$$
$$= \Bigg[\Bigg(\frac{1010\sqrt{2009} + 3018}{2}\Bigg)\Bigg] - 250(\sqrt{2009} - 1)\Bigg]^5$$
$$=(505\sqrt{2009} + 1509 - 250\sqrt{2009} - 250)^5$$
$$= (250\sqrt{2009} - 1259)^5$$
However, the answer given is $32$, so there could have been more simplifications. 
As a question, where did I go wrong? Also can anyone give me some simpler way of solving this?

$a$ is a root of a quadratic equation with roots $$\frac{1 \pm \sqrt{2009}}{2}$$
That is, $a$ satisfies the following equation: $$x^2 - x - 502 = 0 \tag 1$$ Using this, we observe
$$\begin{align}(a^3 - 503a - 500)^5 &= (a(\color{red}{a^2})-503a-500)^5 \\&\overset 1= (a(\color{red}{a+502})-503a-500)^5 \\&= (\color{blue}{a^2-a}-500)^5 \\&\overset 1= (\color{blue}{502} - 500)^5 \\&= 32 \end{align}$$
