Let $\alpha$ and $\beta$ be the roots of $x^2 - 6x - 2 = 0$. If $a_n = \alpha^n - \beta^n$ for $n \ge 1$, then the value of $\frac{a_{10} - 2a_8}{3a_9}$ is:
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If a sequence is defined in terms of the powers of the roots of a quadratic equation $ax^2+bx+c=0$, it will satisfy a linear recurrence relation related to the equation's coefficients, namely $a \cdot u_n + b \cdot u_{n-1} + c \cdot u_{n-2} = 0$.
Since $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 6x - 2 = 0$, they must satisfy the equation.
So, $\alpha^2 - 6\alpha - 2 = 0 \implies \alpha^2 = 6\alpha + 2$.
And, $\beta^2 - 6\beta - 2 = 0 \implies \beta^2 = 6\beta + 2$.
We are given the sequence $a_n = \alpha^n - \beta^n$. We can derive a recurrence relation for this sequence.
Multiply the first equation by $\alpha^{n-2}$ and the second by $\beta^{n-2}$ (for $n \ge 2$):
$\alpha^n = 6\alpha^{n-1} + 2\alpha^{n-2}$.
$\beta^n = 6\beta^{n-1} + 2\beta^{n-2}$.
Subtracting the second equation from the first:
$\alpha^n - \beta^n = 6(\alpha^{n-1} - \beta^{n-1}) + 2(\alpha^{n-2} - \beta^{n-2})$.
Using the definition of $a_n$, this becomes:
$a_n = 6a_{n-1} + 2a_{n-2}$.
We need to evaluate the expression $\frac{a_{10} - 2a_8}{3a_9}$.
Let's apply the recurrence relation for $n=10$:
$a_{10} = 6a_9 + 2a_8$.
Rearranging this equation gives:
$a_{10} - 2a_8 = 6a_9$.
Now substitute this result into the expression we need to find:
$\frac{a_{10} - 2a_8}{3a_9} = \frac{6a_9}{3a_9}$.
Assuming $a_9 \neq 0$, we can cancel the term:
$\frac{6}{3} = 2$.
