We are given three functions $f(x) = x^4 + 3$, $g(x) = x - 6$, and $h(x) = \sqrt{x}$. We need to find $(f \circ g \circ h)(x)$, which means we are looking for the composite function $f(g(h(x)))$.

Let's break it down step by step:

1. First, apply $h(x)$: $h(x) = \sqrt{x}$

2. Next, apply $g(x)$ to $h(x)$: $g(h(x)) = g(\sqrt{x}) = \sqrt{x} - 6$

3. Finally, apply $f(x)$ to $g(h(x))$: $f(g(h(x))) = f(\sqrt{x} - 6)$ Now substitute $\sqrt{x} - 6$ into $f(x) = x^4 + 3$: $f(\sqrt{x} - 6) = (\sqrt{x} - 6)^4 + 3$

Thus, the composite function $(f \circ g \circ h)(x)$ is: $(f \circ g \circ h)(x) = (\sqrt{x} - 6)^4 + 3$

Would you like further details on this process, or any questions related to the steps?

Here are five related questions you might find interesting:

1. What are the steps to find the domain of $f(g(h(x)))$?
2. How would the expression change if $h(x)$ were $x^2$ instead of $\sqrt{x}$?
3. How would you expand $(\sqrt{x} - 6)^4$ using binomial expansion?
4. Can $f(g(h(x)))$ be simplified further in some special cases?
5. What happens to $f(g(h(x)))$ as $x \to \infty$?

Tip: Always evaluate the domain of composite functions to ensure no domain violations occur during substitutions!