1. **State the problem:** Rationalise the expression $$\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}$$ and simplify it. 2. **Formula and rule:** To rationalise a denominator with a sum of square roots, multiply numerator and denominator by the conjugate of the denominator. The conjugate of $$a + b$$ is $$a - b$$. 3. **Apply the conjugate:** Multiply numerator and denominator by $$\sqrt{6} - \sqrt{2}$$: $$\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} \times \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \frac{(\sqrt{6} - \sqrt{2})^2}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})}$$ 4. **Simplify denominator using difference of squares:** $$ (\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4 $$ 5. **Expand numerator:** $$ (\sqrt{6} - \sqrt{2})^2 = (\sqrt{6})^2 - 2 \times \sqrt{6} \times \sqrt{2} + (\sqrt{2})^2 = 6 - 2\sqrt{12} + 2 $$ 6. **Simplify $$\sqrt{12}$$:** $$ \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} $$ 7. **Substitute back:** $$ 6 - 2 \times 2\sqrt{3} + 2 = 6 - 4\sqrt{3} + 2 = 8 - 4\sqrt{3} $$ 8. **Put numerator and denominator together:** $$ \frac{8 - 4\sqrt{3}}{4} = \frac{8}{4} - \frac{4\sqrt{3}}{4} = 2 - \sqrt{3} $$ **Final answer:** $$2 - \sqrt{3}$$ which corresponds to option B.