If the domain of the function \(f(x)=\frac{\sqrt{x^2-25}}{\left(4-x^2\right)}+\log _{10}\left(x^2+2 x-15\right)\) is \((…

We are given:

\(f(x)=\frac{\sqrt{{x}^{2}-25}}{4-{x}^{2}}+{\log }_{10}\left({x}^{2}+2x-15\right)\)

We are told the domain is:

\((-\infty ,\alpha )\cup [\beta ,\infty )\)

We need to find:

\({\alpha }^{2}+{\beta }^{3}\)

Step-by-step (short):
1. Square root condition:

\(\sqrt{{x}^{2}-25}\text{ is defined when }{x}^{2}-25\geq 0\Rightarrow x\leq -5\text{ or }x\geq 5\)

2. Denominator condition:

\(4-{x}^{2}\neq 0\Rightarrow x\neq \pm 2\)

3. Log condition:

\({\log }_{10}\left({x}^{2}+2x-15\right)\text{ defined when }{x}^{2}+2x-15>0\)

Factor:

\((x+5)(x-3)>0\Rightarrow x<-5\text{ or }x>3\)

Now combine all conditions:
- Square root: \(x \leq-5\) or \(x \geq 5\)
- Log: \(x<-5\) or \(x>3\)
- Denominator: \(x \neq \pm 2\)

So:
- For left side: \(x<-5\)

- For right side: Must satisfy \(x \geq 5\) and \(x>3 \Rightarrow x \geq 5\)

Hence, domain is:

\((-\infty ,-5)\cup [5,\infty )\Rightarrow \alpha =-5,\beta =5\)

Now calculate:

\[
\alpha^2+\beta^3=(-5)^2+(5)^3=25+125=150
\]

Final Answer:
(c) 150