Difference of Powers of Roots

We need to simplify the expression $$(11\sqrt{3}-10\sqrt{2})P_{10} + (11\sqrt{2}+10)P_{11} - 11P_{12}.$$

Since $$\alpha, \beta$$ are roots of $$x^2 - \sqrt{2}x - \sqrt{3} = 0$$, we have $$\alpha + \beta = \sqrt{2}$$ and $$\alpha\beta = -\sqrt{3}$$. The sequence $$P_n = \alpha^n - \beta^n$$ therefore satisfies the recurrence $$P_n = \sqrt{2}P_{n-1} + \sqrt{3}P_{n-2}$$, because $$\alpha^n = \sqrt{2}\alpha^{n-1} + \sqrt{3}\alpha^{n-2}$$ (and similarly for $$\beta$$).

In particular, $$P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}$$, so $$11P_{12} = 11\sqrt{2}P_{11} + 11\sqrt{3}P_{10}$$.

Substituting this into the original expression gives

$$(11\sqrt{3}-10\sqrt{2})P_{10} + (11\sqrt{2}+10)P_{11} - \bigl(11\sqrt{2}P_{11} + 11\sqrt{3}P_{10}\bigr).$$

Collecting like terms, the coefficient of $$P_{10}$$ is $$(11\sqrt{3}-10\sqrt{2}) - 11\sqrt{3} = -10\sqrt{2}$$ and the coefficient of $$P_{11}$$ is $$(11\sqrt{2}+10) - 11\sqrt{2} = 10$$. Thus the expression simplifies to $$-10\sqrt{2}P_{10} + 10P_{11}$$.

Factoring out 10 gives $$10(P_{11} - \sqrt{2}P_{10})$$. Using the recurrence backwards, $$P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9$$, so $$P_{11} - \sqrt{2}P_{10} = \sqrt{3}P_9$$ and hence the result is $$10\sqrt{3}P_9$$.

The correct answer is Option 1: $$10\sqrt{3}P_9$$.

We have $$\alpha, \beta$$ as roots of $$x^2 - x - 4 = 0$$ with $$\alpha > \beta$$, and $$P_n = \alpha^n - \beta^n$$. Since $$\alpha$$ and $$\beta$$ satisfy $$x^2 = x + 4$$, we get the recurrence $$P_n = P_{n-1} + 4P_{n-2}$$ for all $$n \geq 2$$.

We simplify the given expression $$\frac{P_{15}P_{16} - P_{14}P_{16} - P_{15}^2 + P_{14}P_{15}}{P_{13}P_{14}}$$. Factoring the numerator: $$P_{16}(P_{15} - P_{14}) - P_{15}(P_{15} - P_{14}) = (P_{15} - P_{14})(P_{16} - P_{15})$$.

From the recurrence $$P_n = P_{n-1} + 4P_{n-2}$$, we get $$P_n - P_{n-1} = 4P_{n-2}$$. Hence $$P_{15} - P_{14} = 4P_{13}$$ and $$P_{16} - P_{15} = 4P_{14}$$.

The expression becomes $$\frac{4P_{13} \cdot 4P_{14}}{P_{13}P_{14}} = \frac{16\,P_{13}P_{14}}{P_{13}P_{14}} = 16$$.

Hence, the correct answer is $$\boxed{16}$$.

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).