1. Problem: Calculate the limit $$\lim_{n \to \infty} \frac{\sqrt[3]{n^2} \sin(n!)}{n - 1}.$$\n\n2. Formula and rules: We analyze the behavior of numerator and denominator as $n \to \infty$.\n- The cube root of $n^2$ is $\sqrt[3]{n^2} = n^{\frac{2}{3}}$.\n- The sine function oscillates between $-1$ and $1$, so $|\sin(n!)| \leq 1$.\n- The denominator $n - 1$ grows without bound as $n \to \infty$.\n\n3. Intermediate work: Rewrite the expression as $$\frac{n^{\frac{2}{3}} \sin(n!)}{n - 1} = \frac{n^{\frac{2}{3}}}{n - 1} \sin(n!).$$\n\n4. Simplify the fraction $$\frac{n^{\frac{2}{3}}}{n - 1} = \frac{n^{\frac{2}{3}}}{n \cancel{- 1}} = \frac{\cancel{n^{\frac{2}{3}}}}{\cancel{n}} n^{-\frac{1}{3}} = n^{-\frac{1}{3}} \text{ as } n \to \infty.$$\n\n5. Since $|\sin(n!)| \leq 1$, the absolute value of the whole expression is bounded by $$\left|\frac{n^{\frac{2}{3}} \sin(n!)}{n - 1}\right| \leq n^{-\frac{1}{3}}.$$\n\n6. As $n \to \infty$, $n^{-\frac{1}{3}} \to 0$, so by the squeeze theorem, the limit is $$0.$$\n\nFinal answer: $$\boxed{0}.$$