To find the domain of the function $$f(x) = \frac{\sqrt{x^2-25}}{(4-x^2)}+\log_{10}(x^2+2x-15),$$ we need to consider the domain conditions for both the square root function and the logarithmic function.

The square root function $\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, so $$x^2 - 25 \geq 0.$$ This inequality is satisfied when $$x \leq -5 \quad \text{or} \quad x \geq 5.$$

The denominator of the rational part of $f(x)$, $(4-x^2)$, cannot be zero, otherwise, the function will become undefined due to division by zero. Thus, we must have $$4 - x^2 \neq 0.$$ This inequality is violated when $$x = \pm2.$$

Combining these conditions gives us the domain for the rational part of the function: $$x \in (-\infty, -5] \cup (5, \infty) \quad \text{and} \quad x \neq 2,-2.$$

Moving on to the logarithmic function, $\log_{10}(x^2+2x-15)$, the argument must be positive: $$x^2 + 2x - 15 > 0.$$ This is a quadratic inequality, which we can factor to find the solution: $$(x+5)(x-3) > 0.$$ From this, we see that the inequality is satisfied for $$x < -5 \quad \text{or} \quad x > 3.$$

The overall domain of $f(x)$ is the intersection of the domains for each piece. Taking the intersection of the two sets gives us: $$x \in (-\infty, -5) \cup (5, \infty),$$

Since the question states that the domain is of the form $(-\infty, \alpha) \cup [\beta, \infty)$, we can infer that $$\alpha = -5 \quad \text{and} \quad \beta = 5.$$

We calculate $\alpha^2 + \beta^3$ as follows: $$\alpha^2 + \beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150.$$

So the correct answer, representing the sum of $\alpha^2$ and $\beta^3$, is: Option D $150$.