In general, for a given recurrence relation of the form $x_n+a_1x_{n-1}+\cdots+a_{r}x_{n-r}=k^nf(n)$, where $a_1,...,a_r,k$ are given constants and $f(n)$ is a polynomial in $n$, and assuming you are given $x_0,\ldots,x_{r-1}$, then you can solve this in two steps:
Step 1: define the characteristic polynomial to be $t^r+a_1t^{r-1}+\cdots+a_0$. Let $t_1,\ldots,t_r$ be all the roots (over $\mathbb{C}$). If they are all distinct, then you write $x_n^{(h)}=\alpha_1t_1^n+\cdots+\alpha_rt_r^n$ - the homogeneous part. If you have multiplicity $s$ to some root $t_j$ then you replace it's appearances in the solution with $(n^{s-1}\beta_1+\cdots+b_{s})t_j^n$.
Step 2: $x_n^{(p)}=k^nn^sg(n)$ where $k$ is the same, $s$ is the multiplicity of $k$ as a root of the characteristic polynomial defined above (if $k$ is not a root, then $s=0$) and $g(n)$ is a polynomial of the same degree as $f(n)$.
Now $x_n=x_n^{(h)}+x_n^{(p)}$. Substituting back and using the given $x_0,\ldots,x_{r-1}$ you can find the constants.
Now, apply this method to your problem: $a_n-a_{n-1}-6a_{n-2}=0$, $a_0=1, a_1=8$.
The characteristic polynomial will be $t^2-t-6=(t-3)(t+2)$. Hence $a_n^{(h)}=\alpha 3^n +\beta (-2)^n$. Since $f(n)=0$ we have $a_n^{(p)}=0$.
Now we just need to use the initial values: $\begin{array}{c} a_0=\alpha+\beta=1 \\ a_1=3\alpha-2\beta=8\end{array} \Rightarrow \begin{array}{c} \alpha+\beta=1 \\ 5\alpha=10\end{array} \Rightarrow \begin{array}{c} \beta=-1 \\ \alpha=2\end{array} $ Finally, $a_n=2\cdot 3^n - (-2)^n$