Answer: B
Let \(x = \sqrt{3} + \sqrt{2}\) Then, \(\frac{1}{x} = \frac{1}{\sqrt{3} + \sqrt{2}}\) \(= \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}\) \(= \sqrt{3} - \sqrt{2}\) Therefore, \(x + \frac{1}{x} = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} = 2\sqrt{3}\) Now, \(x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3x \cdot \frac{1}{x} \cdot (x + \frac{1}{x})\) \(= (2\sqrt{3})^3 - 3 \times 1 \times 2\sqrt{3}\) \(= 24\sqrt{3} - 6\sqrt{3} = 18\sqrt{3}\)
Let \(x = \sqrt{3} + \sqrt{2}\) Then, \(\frac{1}{x} = \frac{1}{\sqrt{3} + \sqrt{2}}\) \(= \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}\) \(= \sqrt{3} - \sqrt{2}\) Therefore, \(x + \frac{1}{x} = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} = 2\sqrt{3}\) Now, \(x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3x \cdot \frac{1}{x} \cdot (x + \frac{1}{x})\) \(= (2\sqrt{3})^3 - 3 \times 1 \times 2\sqrt{3}\) \(= 24\sqrt{3} - 6\sqrt{3} = 18\sqrt{3}\)