If the domain of the function $$f(x) = \frac{\sqrt{x^2 - 25}}{4 - x^2} + \log_{10}(x^2 + 2x - 15)$$ is $$(-\infty, \alpha) \cup [\beta, \infty)$$, then $$\alpha^2 + \beta^3$$ is equal to:

Find $$\alpha^2 + \beta^3$$ for the domain $$(-\infty, \alpha) \cup [\beta, \infty)$$ of $$f(x) = \frac{\sqrt{x^2-25}}{4-x^2} + \log_{10}(x^2+2x-15)$$.

Requires $$x^2 - 25 \geq 0$$, i.e., $$x^2 \geq 25$$.

$$x \leq -5$$ or $$x \geq 5$$.

$$x^2 \neq 4$$, so $$x \neq \pm 2$$.

Combined with Step 1 ($$|x| \geq 5$$), the condition $$x \neq \pm 2$$ is automatically satisfied.

Requires $$x^2 + 2x - 15 > 0$$ (strictly positive for log).

Factoring: $$(x+5)(x-3) > 0$$.

By interval testing: positive when $$x < -5$$ or $$x > 3$$.

From Step 1: $$x \leq -5$$ or $$x \geq 5$$.

From Step 3: $$x < -5$$ or $$x > 3$$.

Intersection: $$(x \leq -5) \cap (x < -5) = x < -5$$, and $$(x \geq 5) \cap (x > 3) = x \geq 5$$.

Wait: at $$x = -5$$: $$x^2 + 2x - 15 = 25 - 10 - 15 = 0$$, and $$\log(0)$$ is undefined. So $$x = -5$$ is excluded.

Domain: $$(-\infty, -5) \cup [5, \infty)$$.

Comparing with $$(-\infty, \alpha) \cup [\beta, \infty)$$: $$\alpha = -5$$ and $$\beta = 5$$.

$$\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150$$.

The correct answer is Option C: 150.