Step 1: Recall the recurrence for $a_n = \alpha^n - \beta^n$.
Since $\alpha, \beta$ satisfy $r^2 = 10r + 8$ (from the quadratic $x^2-10x-8=0$), multiplying by $r^{n-2}$ gives $r^n = 10r^{n-1} + 8r^{n-2}$. This holds for both $\alpha$ and $\beta$.
Step 2: Derive the recurrence for $a_n$.
$a_n = \alpha^n - \beta^n = (10\alpha^{n-1}+8\alpha^{n-2})-(10\beta^{n-1}+8\beta^{n-2}) = 10a_{n-1}+8a_{n-2}$. So: \[ a_n = 10a_{n-1} + 8a_{n-2} \]
Step 3: Apply the recurrence for $n=10$.
\[ a_{10} = 10a_9 + 8a_8 \] Rearranging: \[ a_{10} - 8a_8 = 10a_9 \]
Step 4: Compute the required expression.
\[ \frac{a_{10}-8a_8}{5a_9} = \frac{10a_9}{5a_9} = 2 \] This works as long as $a_9 \neq 0$. Since $\alpha \neq \beta$ (discriminant $= 100+32 = 132 > 0$), $a_9 \neq 0$.
Step 5: Verify the logic is clean.
We never needed explicit values of $a_9$ or $a_8$; the recurrence alone was sufficient. This is the elegant insight of the problem.
Step 6: State the answer.
\[ \boxed{2} \]