Let $$\alpha$$, $$\beta$$ be the roots of the quadratic equation $$x^2 + \sqrt{6}x + 3 = 0$$. Then $$\frac{\alpha^{23} + \beta^{23} + \alpha^{14} + \beta^{14}}{\alpha^{15} + \beta^{15} + \alpha^{10} + \beta^{10}}$$ is equal to

Let $$\alpha, \beta$$ be roots of $$x^2 + \sqrt{6}x + 3 = 0$$. Then $$\alpha + \beta = -\sqrt{6}$$ and $$\alpha\beta = 3$$. Define $$S_n = \alpha^n + \beta^n$$, which satisfies the recurrence $$S_n = (\alpha+\beta)S_{n-1} - \alpha\beta S_{n-2} = -\sqrt{6}S_{n-1} - 3S_{n-2}$$.

We have $$S_0 = 2$$
$$S_1 = -\sqrt{6}$$
$$S_2 = (\alpha+\beta)^2 - 2\alpha\beta = 6 - 6 = 0$$
$$S_3 = -\sqrt{6}(0) - 3(-\sqrt{6}) = 3\sqrt{6}$$
$$S_4 = -\sqrt{6}(3\sqrt{6}) - 3(0) = -18$$
$$S_5 = -\sqrt{6}(-18) - 3(3\sqrt{6}) = 18\sqrt{6} - 9\sqrt{6} = 9\sqrt{6}$$
$$S_6 = -\sqrt{6}(9\sqrt{6}) - 3(-18) = -54 + 54 = 0$$

Since $$S_2 = 0$$ and $$S_6 = 0$$ and in general $$S_{4k+2} = 0$$ for all $$k \geq 0$$, it follows that $$S_{10} = 0$$ and $$S_{14} = 0$$.

Therefore, the given expression simplifies as $$\frac{S_{23} + S_{14}}{S_{15} + S_{10}} = \frac{S_{23} + 0}{S_{15} + 0} = \frac{S_{23}}{S_{15}}$$, and from the recurrence one finds $$S_{n+8} = 81 S_n$$ so that $$\frac{S_{23}}{S_{15}} = \frac{S_{15+8}}{S_{15}} = 81$$.

Alternatively, one may note that $$S_{23} = -\sqrt{6} S_{22} - 3 S_{21}$$ and continue the pattern with $$S_{15}$$ as reference to obtain the same ratio.

81