1. We need the domain of $f(x)=\frac{\sqrt{x^2-25}}{4-x^2}+\log_{10}(x^2+2x-15).$

The domain will be the intersection of the domains of both parts.

2. Condition from the square root: $x^2-25\ge 0$ $x^2\ge 25$ $|x|\ge 5$ So, $x\le -5 \quad \text{or} \quad x\ge 5.$

3. Condition from the denominator: $4-x^2\ne 0$ $x^2\ne 4$ $x\ne \pm 2.$ But since from the square root condition we already have $x\le -5$ or $x\ge 5$, the values $\pm 2$ do not lie there. So this gives no extra restriction.

4. Condition from the logarithm: $x^2+2x-15>0$ Factorizing, $x^2+2x-15=(x+5)(x-3)>0.$ This is positive when $x<-5 \quad \text{or} \quad x>3.$

5. Intersect all conditions:

• From square root: $(-\infty,-5]\cup[5,\infty)$
• From logarithm: $(-\infty,-5)\cup(3,\infty)$

Their intersection is $(-\infty,-5)\cup[5,\infty).$

So comparing with $(-\infty,\alpha)\cup[\beta,\infty),$ we get $\alpha=-5,\qquad \beta=5.$

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

7. Option check:

• A: $140$
• B: $175$
• C: $125$
• D: $150$ ✅

Hence the correct answer is D.