To find the domain of the function $g(x) = \frac{3}{\sqrt[3]{x^2 - 3x + 2}}$, we need to consider the properties of the cube root and the denominator:

1. Cube Root: The cube root function, $\sqrt[3]{x}$, is defined for all real numbers, meaning that the expression inside the cube root doesn't impose any restrictions on the domain by itself.

2. Denominator: However, we must ensure that the denominator is not equal to zero because division by zero is undefined. Therefore, we need to solve: $x^2 - 3x + 2 = 0$ Factorizing the quadratic equation: $(x - 1)(x - 2) = 0$ So, $x = 1$ and $x = 2$ are the values that make the denominator zero, which means $x = 1$ and $x = 2$ must be excluded from the domain.

Conclusion:

The domain of the function is all real numbers except $x = 1$ and $x = 2$, meaning the domain is: $(-\infty, 1) \cup (1, 2) \cup (2, +\infty)$ This corresponds to option D.

Would you like further details on this? Here are some additional questions to expand the understanding:

1. How would the domain change if it were a square root instead of a cube root?
2. Can you describe why $x = 1$ and $x = 2$ are excluded from the domain?
3. What happens to the function's graph around $x = 1$ and $x = 2$?
4. How do cube roots differ from even roots in terms of domain restrictions?
5. Could there be any vertical asymptotes for this function?

Tip: Always factor the denominator first to identify critical values that must be excluded from the domain.