The given function is:

$w = h(u) = \sqrt[3]{u - 3}$

To determine the domain of this function, we need to find the set of values for $u$ for which the function is defined.

Since this is a cube root function, we know that cube roots are defined for all real numbers. There are no restrictions on the domain from the cube root operation, unlike square roots which require non-negative numbers inside the radical.

Thus, the domain of $h(u)$ is:

$\boxed{(-\infty, \infty)}$

Would you like more details or have any further questions?

Related Questions:

1. What is the range of the function $h(u) = \sqrt[3]{u - 3}$?
2. How would the domain change if this were a square root instead of a cube root?
3. How do you graph the function $h(u) = \sqrt[3]{u - 3}$?
4. What is the derivative of $h(u) = \sqrt[3]{u - 3}$?
5. How does shifting the function to $h(u) = \sqrt[3]{u - 3} + 2$ affect the domain?

Tip: For cube root functions, there are no domain restrictions, but always check for domain issues in even-root functions (like square roots).