1. **State the problem:** Rationalise the denominator of the expression $$\frac{2\sqrt{2}}{\sqrt{5} - \sqrt{3}}$$ and express the answer in the form $$a\sqrt{10} + b\sqrt{6}$$. 2. **Formula and rule:** To rationalise a denominator with a difference of square roots, multiply numerator and denominator by the conjugate of the denominator. The conjugate of $$\sqrt{5} - \sqrt{3}$$ is $$\sqrt{5} + \sqrt{3}$$. 3. **Multiply numerator and denominator by the conjugate:** $$\frac{2\sqrt{2}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{2\sqrt{2}(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}$$ 4. **Simplify the denominator using difference of squares:** $$ (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2 $$ 5. **Expand the numerator:** $$ 2\sqrt{2} \times \sqrt{5} + 2\sqrt{2} \times \sqrt{3} = 2\sqrt{10} + 2\sqrt{6} $$ 6. **Write the fraction:** $$ \frac{2\sqrt{10} + 2\sqrt{6}}{2} $$ 7. **Simplify by dividing numerator and denominator by 2:** $$ \frac{\cancel{2}\sqrt{10} + \cancel{2}\sqrt{6}}{\cancel{2}} = \sqrt{10} + \sqrt{6} $$ **Final answer:** $$\sqrt{10} + \sqrt{6}$$