If $a = \frac{1+\sqrt5}2$, then what is $a^{18} + \frac{323}{a^6}$? My question is this:

If $a = \frac{1+\sqrt5}2,\frac{1-\sqrt5}2$, then what is $a^{18} + \frac{323}{a^6}$?

It is an AMC style question and is timed, so I will not be able to use solutions with a lot of case work.
I defined $a$ from part a of a 2 part question and would like to know how to do this without the use of a calculator.


Lemma.  Define $$L_n = a^n + (-a)^{-n}.$$  Then $L_n$ is a Lucas number satisfying the recurrence $$L_{n+1} = L_n + L_{n-1}, \\ L_0 = 2, \quad L_1 = 1.$$
Proof.  Since $a^2 = a + 1$, or equivalently, $$a - a^{-1} = 1,$$ we have $$(a^n + (-a)^{-n})(a - a^{-1}) = a^{n+1} + (-a)^{-(n+1)} - \left(a^{n-1} + (-a)^{-(n+1)}\right),$$ hence $$L_n L_1 = L_{n+1} - L_{n-1}.$$
Along with $L_0 = a^0 + (-a)^0 = 2$, this proves the claim.

We next use this recursion to compute the table
$$\begin{array}{c|ccccccccccccc}
n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
L_n & 2 & 1 & 3 & 4 & 7 & 11 & 18 & 29 & 47 & 76 & 123 & 199 & 322 
\end{array}$$
Therefore
$$\begin{align}
a^{18} + 323 a^{-6} 
&= a^6 (a^{12} + 323a^{-12}) \\
&= a^6(a^{12} + a^{-12} + 322a^{-12}) \\
&= a^6 (L_{12} + 322 a^{-12}) \\
&= a^6 (322 + 322 a^{-12}) \\
&= 322 (a^6 + a^{-6}) \\
&= 322 L_6 \\
&= 322 (18) \\
&= 5796.
\end{align}$$
