1. The problem is to simplify the expression $$\frac{12}{\sqrt{7} - \sqrt{3}} + \frac{12}{\sqrt{7} + \sqrt{3}}$$. 2. To simplify expressions with square roots in the denominator, we use the conjugate to rationalize the denominator. The conjugate of $\sqrt{7} - \sqrt{3}$ is $\sqrt{7} + \sqrt{3}$ and vice versa. 3. Rationalize each fraction: $$\frac{12}{\sqrt{7} - \sqrt{3}} \times \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} + \sqrt{3}} = \frac{12(\sqrt{7} + \sqrt{3})}{(\sqrt{7})^2 - (\sqrt{3})^2} = \frac{12(\sqrt{7} + \sqrt{3})}{7 - 3} = \frac{12(\sqrt{7} + \sqrt{3})}{4} = 3(\sqrt{7} + \sqrt{3})$$ $$\frac{12}{\sqrt{7} + \sqrt{3}} \times \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} - \sqrt{3}} = \frac{12(\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{12(\sqrt{7} - \sqrt{3})}{4} = 3(\sqrt{7} - \sqrt{3})$$ 4. Add the two results: $$3(\sqrt{7} + \sqrt{3}) + 3(\sqrt{7} - \sqrt{3}) = 3\sqrt{7} + 3\sqrt{3} + 3\sqrt{7} - 3\sqrt{3} = 3\sqrt{7} + 3\sqrt{7} = 6\sqrt{7}$$ 5. The simplified expression is $$6\sqrt{7}$$. This means the original expression simplifies neatly to six times the square root of seven.