Example 5 (2005 Jiangsu Winter Camp) Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{0}=1, b_{0}=0$,
and $\left\{\begin{array}{l}a_{n+1}=7 a_{n}+6 b_{n}-3, \\ b_{n+1}=8 a_{n}+7 b_{n}-4 .\end{array}\right.$
(1)
(2) $n=0,1,2, \cdots$
Find $a_{n}$.
Given $a_{0}=1, b_{0}=0$, we get $a_{1}=4, b_{1}=4, a_{2}=49$.
(1) $\times 7: 7 a_{n+1}=49 a_{n}+42 b_{n}-21$,
(2) $\times 6: 6 b_{n+1}=48 a_{n}+42 b_{n}-24$.
Subtracting the two equations, we get $6 b_{n+1}-7 a_{n+1}=-a_{n}-3$,
which simplifies to $6 b_{n}=7 a_{n}-a_{n-1}-3$.
Substituting into (1): $a_{n+1}=14 a_{n}-a_{n-1}-6$.
Thus, $a_{n+1}-\frac{1}{2}=14\left(a_{n}-\frac{1}{2}\right)-\left(a_{n-1}-\frac{1}{2}\right)$.
The characteristic equation is $x^{2}-14 x+1=0$, with solutions $x=7 \pm 4 \sqrt{3}$.

Therefore, $a_{n}=\alpha(7+4 \sqrt{3})^{n}+\beta(7-4 \sqrt{3})^{n}+\frac{1}{2}$. Given $a_{0}=1, a_{1}=4, a_{2}=49$, we solve to get $\alpha=\beta=\frac{1}{4}$.
Thus, $a_{n}=\frac{1}{4}(7+4 \sqrt{3})^{n}+\frac{1}{4}(7-4 \sqrt{3})^{n}+\frac{1}{2}$
$$
\begin{array}{l}
=\frac{1}{4}(2+\sqrt{3})^{2 n}+\frac{1}{4}(2-\sqrt{3})^{2 n}+\frac{1}{2}= \\
{\left[\frac{1}{2}(2+\sqrt{3})^{n}+\frac{1}{2}(2-\sqrt{3})^{n}\right]^{2}}
\end{array}
$$