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.