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(C) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 6x - 2 = 0$.From the properties of roots,we have $\alpha + \beta =

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$3$

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(C) Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 6x - 2 = 0$.From the properties of roots,we have $\alpha + \beta = 6$ and $\alpha \beta = -2$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$\alpha^2 - 6\alpha - 2 = 0 \implies \alpha^2 - 2 = 6\alpha$$\beta^2 - 6\beta - 2 = 0 \implies \beta^2 - 2 = 6\beta$We are given $a_n = \alpha^n - \beta^n$. We need to evaluate $\frac{a_{10} - 2a_8}{2a_9}$.Substituting the expression for $a_n$:$\frac{a_{10} - 2a_8}{2a_9} = \frac{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)}{2(\alpha^9 - \beta^9)}$$= \frac{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)}{2(\alpha^9 - \beta^9)}$Using the relations $\alpha^2 - 2 = 6\alpha$ and $\beta^2 - 2 = 6\beta$:$= \frac{\alpha^8(6\alpha) - \beta^8(6\beta)}{2(\alpha^9 - \beta^9)}$$= \frac{6\alpha^9 - 6\beta^9}{2(\alpha^9 - \beta^9)}$$= \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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