1. Stating the problem: We want to find the limit $$\lim_{n \to \infty} \sqrt{n^3 \left( \sqrt{n^3 + 1} - \sqrt{n^3 - 2} \right)}.$$\n\n2. Simplify the expression inside the parentheses using the difference of squares technique: multiply and divide by the conjugate \(\sqrt{n^3 + 1} + \sqrt{n^3 - 2}\) to rationalize it.\n\n3. This gives:\n$$\sqrt{n^3 + 1} - \sqrt{n^3 - 2} = \frac{(n^3 + 1) - (n^3 - 2)}{\sqrt{n^3 + 1} + \sqrt{n^3 - 2}} = \frac{3}{\sqrt{n^3 + 1} + \sqrt{n^3 - 2}}.$$\n\n4. Substitute back into the original limit:\n$$\lim_{n \to \infty} \sqrt{n^3 \cdot \frac{3}{\sqrt{n^3 + 1} + \sqrt{n^3 - 2}}} = \lim_{n \to \infty} \sqrt{\frac{3 n^3}{\sqrt{n^3 + 1} + \sqrt{n^3 - 2}}}.$$\n\n5. For large \(n\), \(\sqrt{n^3 + 1} \approx \sqrt{n^3} = n^{3/2}\) and \(\sqrt{n^3 - 2} \approx n^{3/2}\). So the denominator \(\approx n^{3/2} + n^{3/2} = 2 n^{3/2}\).\n\n6. Substitute this approximation:\n$$\lim_{n \to \infty} \sqrt{\frac{3 n^3}{2 n^{3/2}}} = \lim_{n \to \infty} \sqrt{\frac{3}{2} n^{3 - \frac{3}{2}}} = \lim_{n \to \infty} \sqrt{\frac{3}{2} n^{\frac{3}{2}}}.$$\n\n7. Rewrite the square root:\n$$\sqrt{\frac{3}{2} n^{\frac{3}{2}}} = \sqrt{\frac{3}{2}} \cdot \sqrt{n^{\frac{3}{2}}} = \sqrt{\frac{3}{2}} \cdot n^{\frac{3}{4}}.$$\n\n8. As \(n \to \infty\), \(n^{3/4} \to \infty\), so the limit diverges to infinity.\n\nFinal answer: $$\lim_{n \to \infty} \sqrt{n^3 (\sqrt{n^3 + 1} - \sqrt{n^3 - 2})} = \infty.$$