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We are given the following limit expression:

\[
\lim_{n \to \infty} \sqrt[3]{n} \cdot \left( \sqrt[3]{n^2} - \sqrt[3]{n(n-1)} \right)
\]

### Step-by-Step Solution

1. **Rewrite the expression:**
   Let's simplify the cube roots:

   - \(\sqrt[3]{n} = n^{1/3}\)
   - \(\sqrt[3]{n^2} = n^{2/3}\)
   - \(\sqrt[3]{n(n-1)} = \sqrt[3]{n^2 - n} = n^{2/3} \cdot \sqrt[3]{1 - \frac{1}{n}}\)

   Substituting these into the limit expression:

   \[
   \lim_{n \to \infty} n^{1/3} \left( n^{2/3} - n^{2/3} \cdot \sqrt[3]{1 - \frac{1}{n}} \right)
   \]

2. **Factor out \(n^{2/3}\):**

   \[
   \lim_{n \to \infty} n^{1/3} \cdot n^{2/3} \left( 1 - \sqrt[3]{1 - \frac{1}{n}} \right)
   \]

   This simplifies further to:

   \[
   \lim_{n \to \infty} n \left( 1 - \sqrt[3]{1 - \frac{1}{n}} \right)
   \]

3. **Expand the cube root using a series approximation:**

   For large \(n\), \(\sqrt[3]{1 - \frac{1}{n}}\) can be approximated by a Taylor expansion:

   \[
   \sqrt[3]{1 - \frac{1}{n}} \approx 1 - \frac{1}{3n} + O\left(\frac{1}{n^2}\right)
   \]

   Substituting this approximation into the expression:

   \[
   \lim_{n \to \infty} n \left( 1 - \left(1 - \frac{1}{3n}\right) \right)
   \]

   Simplifying:

   \[
   \lim_{n \to \infty} n \cdot \frac{1}{3n} = \frac{1}{3}
   \]

Thus, the value of the limit is:

\[
\boxed{\frac{1}{3}}
\]

Let me know if you would like more details or further explanations! Here are some related questions:

1. How do Taylor expansions help in simplifying limits involving roots?
2. What other approximations are useful for limits as \(n \to \infty\)?
3. How does the cube root function behave near infinity?
4. Can this method be applied to higher-order roots or powers?
5. How would you handle limits involving different types of logarithms?

**Tip:** When approaching limits, approximating complicated expressions with series expansions can significantly simplify the evaluation.

Evaluate the limit: lim (n -> ∞) (cube root of n) * (cube root of n^2 - cube root of n(n-1)).

This solution walks through the limit evaluation lim (n -> ∞) (cube root of n) * (cube root of n^2 - cube root of n(n-1)). Using algebraic simplifications and a Taylor series approximation, the final answer is found to be 1/3. Suitable for college-level calculus students.

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