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(C) Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$,we have $\alpha + \beta = 6$ and $\alpha\beta = -2$.Also,$\alpha^2 = 6\alpha + 2$ and $

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$3$

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(C) Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$,we have $\alpha + \beta = 6$ and $\alpha\beta = -2$.Also,$\alpha^2 = 6\alpha + 2$ and $\beta^2 = 6\beta + 2$.Given $a_n = \alpha^n - \beta^n$,we evaluate the expression:$\frac{a_{10} - 2a_8}{2a_9} = \frac{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)}{2(\alpha^9 - \beta^9)}$$= \frac{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)}{2(\alpha^9 - \beta^9)}$Since $\alpha^2 - 2 = 6\alpha$ and $\beta^2 - 2 = 6\beta$,we substitute these values:$= \frac{\alpha^8(6\alpha) - \beta^8(6\beta)}{2(\alpha^9 - \beta^9)}$$= \frac{6(\alpha^9 - \beta^9)}{2(\alpha^9 - \beta^9)}$$= \frac{6}{2} = 3$.

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - 6x - 2 = 0$. If $a_n = \alpha^n - \beta^n$ for $n \ge 1$,then the value of $\frac{a_{10} - 2a_8}{2a_9}$ is equal to:

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