Answer: B
Here's the English translation of your solution: Let \(x + y = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} = 2\sqrt{3}\) \(xy = (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1\) Therefore, \(8xy(x^2 + y^2)\) \(= 8xy[(x + y)^2 - 2xy]\) \(= 8 \times 1 \times [(2\sqrt{3})^2 - 2 \times 1]\) \(= 8[12 - 2] = 8 \times 10 = 80\)
Here's the English translation of your solution: Let \(x + y = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} = 2\sqrt{3}\) \(xy = (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1\) Therefore, \(8xy(x^2 + y^2)\) \(= 8xy[(x + y)^2 - 2xy]\) \(= 8 \times 1 \times [(2\sqrt{3})^2 - 2 \times 1]\) \(= 8[12 - 2] = 8 \times 10 = 80\)