The domain of $f\left(x\right)=\log \left(\frac{x-5}{x^2-10x+24}\right)-\sqrt[3]{x+5}$  is

Domain of $\sqrt[3]{x+5}$ is $R$

$x-5>0$ and $x^2-10x+24>0$

$x>5$ and$\left(x-4\right)\left(x-6\right)>0$

$x>5$ and $x<4$ or $x>6$

$\therefore x\in \left(6,\infty \right)$

Case(ii):- $x-5<0$ and $x^2-10x+24<0$

$x<5$ and $\left(x-4\right)\left(x-6\right)<0$

$x<5$ and $4<x<6$ $\therefore x\in \left(4,5\right)$

Hence domain of $f\left(x\right)$ is $\left(4,5\right)\cup \left(6,\infty \right)$