To find the domain of the function, we need to identify any values of $x$ that would result in division by zero or the square root of a negative number, as these are not defined.
Let's look at each term separately.
The first term, $\frac{1}{x+8}$, is defined for all values of $x$ except when the denominator, $x+8$, equals zero. So, $x+8 \neq 0 \implies x \neq -8$.
The second term, $\frac{1}{\sqrt{x - 8}}$, is defined for all values of $x$ greater than or equal to 8, as the square root of a negative number is not defined for real numbers.
The third term, $\frac{1}{\sqrt{8 - x}}$, is defined for all values of $x$ less than or equal to 8, as the square root of a negative number is not defined for real numbers.
Combining these conditions, we find that the domain of the function is the set of all values of $x$ that satisfy the following conditions:
$x \neq -8$
$x \geq 8$
$x \leq 8$
This can be expressed as a union of intervals. Breaking down the intervals:
$x \neq -8$: This can be represented by $(-\infty, -8) \cup (-8, \infty)$.
$x \geq 8$: This can be represented by $[8, \infty)$.
$x \leq 8$: This can be represented by $(-\infty, 8]$.
Therefore, the domain of the function is $(-\infty, -8) \cup (-8, 8] \cup [8, \infty)$.