Let $$\alpha_1, \alpha_2, \ldots, \alpha_7$$ be the roots of the equation $$x^7 + 3x^5 - 13x^3 - 15x = 0$$ and $$|\alpha_1| \geq |\alpha_2| \geq \ldots \geq |\alpha_7|$$. Then, $$\alpha_1\alpha_2 - \alpha_3\alpha_4 + \alpha_5\alpha_6$$ is equal to ______.

Find $$\alpha_1\alpha_2 - \alpha_3\alpha_4 + \alpha_5\alpha_6$$ where $$\alpha_1, \ldots, \alpha_7$$ are roots of $$x^7 + 3x^5 - 13x^3 - 15x = 0$$ ordered by decreasing absolute value.

We factor the equation as $$x(x^6 + 3x^4 - 13x^2 - 15) = 0$$ and, letting $$u = x^2$$, obtain $$u^3 + 3u^2 - 13u - 15 = 0$$. Testing $$u = 3$$ gives $$27 + 27 - 39 - 15 = 0$$, so the cubic factors as $$(u-3)(u^2 + 6u + 5) = (u-3)(u+1)(u+5) = 0$$, yielding $$u = 3, -1, -5$$.

Therefore, $$x^2 = 3$$ implies $$x = \pm\sqrt{3}$$ (so $$|\alpha| = \sqrt{3}$$); $$x^2 = -1$$ implies $$x = \pm i$$ (so $$|\alpha| = 1$$); $$x^2 = -5$$ implies $$x = \pm i\sqrt{5}$$ (so $$|\alpha| = \sqrt{5}$$); and also $$x = 0$$ (so $$|\alpha| = 0$$).

Ordering the roots by decreasing absolute value gives $$|\alpha_1| = |\alpha_2| = \sqrt{5}$$, $$|\alpha_3| = |\alpha_4| = \sqrt{3}$$, $$|\alpha_5| = |\alpha_6| = 1$$, and $$|\alpha_7| = 0$$.

It follows that $$\alpha_1\alpha_2 = (i\sqrt{5})(-i\sqrt{5}) = 5$$, $$\alpha_3\alpha_4 = (\sqrt{3})(-\sqrt{3}) = -3$$, and $$\alpha_5\alpha_6 = (i)(-i) = 1$$.

Hence $$\alpha_1\alpha_2 - \alpha_3\alpha_4 + \alpha_5\alpha_6 = 5 - (-3) + 1 = 9$$. The answer is $$\boxed{9}$$.