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The problem is to evaluate the limit:

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

First, notice that \(\sqrt[3]{n^2} = n^{2/3}\) and \(\sqrt[3]{n(n-1)}\) can be approximated as \(\sqrt[3]{n^2 - n} = (n^2 - n)^{1/3}\). As \(n\) becomes very large, the term \(-n\) inside the cube root becomes insignificant compared to \(n^2\), so \(\sqrt[3]{n(n-1)} \approx n^{2/3}\). Therefore, we can simplify the expression:

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

Now, substituting this back into the original limit:

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

Thus, the limit diverges to \(-\infty\).

Would you like further details on this problem or any related concept? Here are some additional questions you could consider:

1. What happens if we do not approximate and calculate \(\sqrt[3]{n(n-1)}\) exactly?
2. How would the limit change if the constant \(3\) in the expression was a different constant?
3. Can this method be used for limits involving other functions with \(n^{k}\) where \(k\) is not \(2/3\)?
4. What are other techniques to evaluate limits involving cube roots?
5. How does the behavior of \(\sqrt[3]{n(n-1)}\) compare to \(\sqrt[3]{n^2}\) as \(n\) approaches infinity?

**Tip:** When evaluating limits of sequences involving roots, consider the leading terms as \(n\) grows, as they will have the most significant impact on the limit's value.

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

This detailed solution evaluates the limit involving cube roots and large values of n. Learn how to simplify expressions and approximate terms when dealing with limits involving cube roots. Suitable for undergraduate calculus students.

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