Let $$\alpha$$ and $$\beta$$ be the roots of $$x^2 - 6x - 2 = 0$$. If $$a_n = \alpha^n - \beta^n$$ for $$n \geq 1$$, then the value of $$\dfrac{a_{10} - 2a_8}{3a_9}$$ is:

Given that $$\alpha$$ and $$\beta$$ are the roots of $$x^2 - 6x - 2 = 0$$, by Vieta's formulas we have $$\alpha + \beta = 6$$ and $$\alpha\beta = -2$$. Also, since both $$\alpha$$ and $$\beta$$ satisfy the equation, we have $$\alpha^2 = 6\alpha + 2$$ and $$\beta^2 = 6\beta + 2$$.

We need to find $$\frac{a_{10} - 2a_8}{3a_9}$$ where $$a_n = \alpha^n - \beta^n$$.

Consider $$a_{10} - 2a_8 = (\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8) = \alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)$$.

Using $$\alpha^2 = 6\alpha + 2$$, we get $$\alpha^2 - 2 = 6\alpha$$. Similarly, $$\beta^2 - 2 = 6\beta$$.

Substituting: $$a_{10} - 2a_8 = \alpha^8(6\alpha) - \beta^8(6\beta) = 6(\alpha^9 - \beta^9) = 6a_9$$.

Therefore, $$\frac{a_{10} - 2a_8}{3a_9} = \frac{6a_9}{3a_9} = 2$$.

The answer is 2, which corresponds to option (3).