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

$\Rightarrow x^2-25\ge 0$

$\Rightarrow \left(x-5\right)\left(x+5\right)\ge 0$

$\Rightarrow x\in (-\infty ,-5]\cup [5,\infty ) ...\left(i\right)$

$4-x^2\ne 0$

$\Rightarrow x\ne 2,-2 ...\left(ii\right)$

$\Rightarrow x^2+2x-15>0$

$\Rightarrow \left(x+5\right)\left(x-3\right)>0$

$\Rightarrow x\in \left(-\infty ,-5\right)\cup \left(3,\infty \right) ...\left(iii\right)$

Using $\left(i\right),\left(ii\right) \text{and} \left(iii\right)$

$\Rightarrow x\in \left(-\infty ,-5\right)\cup [5,\infty )$

So, on comparing with $x\in \left(-\infty ,\alpha \right)\cup [\beta ,\infty )$ we get,

$\Rightarrow {\alpha }^2+{\beta }^3={\left(-5\right)}^2+{\left(5\right)}^3$

$\Rightarrow {\alpha }^2+{\beta }^3=150$