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: $a$ 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
$$\begin{align}(a^3 - 503a - 500)^5 &= (a(\color{red}{a^2})-503a-500)^5 \\&\overset 1= (a(\color{red}{a+502})-503a-500)^5 \\&= (\color{blue}{a^2-a}-500)^5 \\&\overset 1= (\color{blue}{502} - 500)^5 \\&= 32 \end{align}$$
A: Your working leads to the answer. Here is the correction -
$a = \frac{1 + \sqrt{2009}}{2}$
$a^2 - 3 = \frac{999 + \sqrt{2009}}{2}$
$a (a^2 - 3) = 752 + 250 \sqrt {2009}$
$500 (a + 1) = 250 ( \sqrt {2009} + 3)$
$a(a^2-3) - 500(a+1) = 2$
So $(a^3 - 503a - 500)^5 = 32$
A: 
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .

I see no reason for elegance.  Since one of the factors in the numerator is $1$, computing $a^3$ is is straightforward.
$$a^3 = \left(\frac{1}{8}\right) \times 
\left[
1 + 3\sqrt{2009} + 3(2009) + 2009\sqrt{2009}
\right]$$
$$=~ 
\left(\frac{1}{8}\right) \times
\left[6028 + 2012\sqrt{2009}\right]
~=~
\frac{1507 + 503\sqrt{2009}}{2}.
$$
Therefore,
$$(a^3 - 503a - 500)$$
$$=~ \frac{1507 + 503\sqrt{2009}}{2} ~-~ 
\frac{503 + 503\sqrt{2009}}{2} - \frac{1000}{2}
~=~ \frac{4}{2} \implies $$
$$(a^3 - 503a - 500)^5 = 2^5 = 32.$$
