Find all real numbers $x$ for which $$\sqrt{\frac{x^3 - 8}{x}} > x - 2.$$
1. **Identify the domain constraints:**
   - The expression inside the square root, $\frac{x^3 - 8}{x}$, must be non-negative:
     \[
     \frac{x^3 - 8}{x} \geq 0
     \]
     This implies:
     \[
     x^3 - 8 \geq 0 \quad \text{and} \quad x \neq 0
     \]
     \[
     (x-2)(x^2 + 2x + 4) \geq 0
     \]
     Since $x^2 + 2x + 4 > 0$ for all $x$, we only need:
     \[
     x - 2 \geq 0 \implies x \geq 2 \quad \text{or} \quad x < 0
     \]

2. **Consider the inequality $\sqrt{\frac{x^3 - 8}{x}} > x - 2$:**
   - For $x \geq 2$:
     \[
     \sqrt{\frac{x^3 - 8}{x}} > x - 2
     \]
     Squaring both sides (since both sides are non-negative for $x \geq 2$):
     \[
     \frac{x^3 - 8}{x} > (x - 2)^2
     \]
     Simplifying:
     \[
     \frac{x^3 - 8}{x} > x^2 - 4x + 4
     \]
     \[
     x^3 - 8 > x(x^2 - 4x + 4)
     \]
     \[
     x^3 - 8 > x^3 - 4x^2 + 4x
     \]
     \[
     -8 > -4x^2 + 4x
     \]
     \[
     4x^2 - 4x - 8 > 0
     \]
     \[
     x^2 - x - 2 > 0
     \]
     Factoring:
     \[
     (x - 2)(x + 1) > 0
     \]
     This inequality holds for:
     \[
     x > 2 \quad \text{or} \quad x < -1
     \]

   - For $x < 0$:
     \[
     \sqrt{\frac{x^3 - 8}{x}} > x - 2
     \]
     Since $\frac{x^3 - 8}{x}$ is positive and $x - 2$ is negative, the inequality holds for all $x < 0$.

3. **Combine the intervals:**
   - From the analysis, the solution is:
     \[
     x \in (-\infty, 0) \cup (2, \infty)
     \]

The final answer is $\boxed{(-\infty, 0) \cup (2, \infty)}$