The limit given is:

\[
\lim_{h \to 0} \frac{(x-h)^3 - x^3}{h}
\]

### Steps to Evaluate the Limit:

1. **Expand \((x - h)^3\) using the binomial theorem**:

   \[
   (x - h)^3 = x^3 - 3x^2h + 3xh^2 - h^3
   \]

2. **Substitute this expansion into the limit expression**:

   \[
   \lim_{h \to 0} \frac{x^3 - 3x^2h + 3xh^2 - h^3 - x^3}{h}
   \]

3. **Simplify the expression**:
   
   Notice that \(x^3\) and \(-x^3\) cancel out:

   \[
   \lim_{h \to 0} \frac{-3x^2h + 3xh^2 - h^3}{h}
   \]

   Factor \(h\) out of the numerator:

   \[
   \lim_{h \to 0} \frac{h(-3x^2 + 3xh - h^2)}{h}
   \]

   Cancel \(h\) in the numerator and denominator:

   \[
   \lim_{h \to 0} (-3x^2 + 3xh - h^2)
   \]

4. **Evaluate the limit as \(h\) approaches 0**:
   
   As \(h \to 0\), the terms involving \(h\) go to 0:

   \[
   -3x^2 + 3x(0) - 0^2 = -3x^2
   \]

So, the limit is:

\[
\boxed{-3x^2}
\]

Would you like more details on this process or have any other questions?

### Related Questions:

1. How can this result be connected to the derivative of a function?
2. What would the limit be if the function was \((x-h)^2 - x^2\) instead of \((x-h)^3 - x^3\)?
3. How can this limit be derived using the definition of the derivative?
4. What would happen if the expression involved higher powers, like \((x-h)^n - x^n\)?
5. How does this approach differ from evaluating a limit using L'Hôpital's Rule?

### Tip:
The limit you've just calculated is a common derivative form. Specifically, it represents the derivative of \(x^3\) at a point \(x\).

Learn how to evaluate the limit of (x-h)^3 - x^3 / h using the binomial theorem and differentiation techniques. This problem involves understanding limits and their connection to differentiation in calculus. Suitable for advanced high school students and college-level calculus enthusiasts.

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