1. **Problem Statement:** Rationalise the denominators of the given expressions and simplify where possible. 2. **Formula and Rule:** To rationalise a denominator containing a square root, multiply numerator and denominator by the conjugate of the denominator. The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$ and vice versa. 3. **Solutions:** **(a) $\frac{6\sqrt{3}}{\sqrt{8} - \sqrt{7}}$** Multiply numerator and denominator by the conjugate $\sqrt{8} + \sqrt{7}$: $$\frac{6\sqrt{3}}{\sqrt{8} - \sqrt{7}} \times \frac{\sqrt{8} + \sqrt{7}}{\sqrt{8} + \sqrt{7}} = \frac{6\sqrt{3}(\sqrt{8} + \sqrt{7})}{(\sqrt{8})^2 - (\sqrt{7})^2}$$ Calculate denominator: $$8 - 7 = 1$$ So, $$= 6\sqrt{3}(\sqrt{8} + \sqrt{7}) = 6\sqrt{3}\sqrt{8} + 6\sqrt{3}\sqrt{7}$$ Simplify radicals: $$\sqrt{8} = 2\sqrt{2}$$ So, $$= 6\sqrt{3} \times 2\sqrt{2} + 6\sqrt{3}\sqrt{7} = 12\sqrt{6} + 6\sqrt{21}$$ **Answer:** $12\sqrt{6} + 6\sqrt{21}$ **(b) $\frac{\sqrt{7}}{\sqrt{7} - \sqrt{3}}$** Multiply numerator and denominator by conjugate $\sqrt{7} + \sqrt{3}$: $$\frac{\sqrt{7}}{\sqrt{7} - \sqrt{3}} \times \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} + \sqrt{3}} = \frac{\sqrt{7}(\sqrt{7} + \sqrt{3})}{7 - 3}$$ Calculate denominator: $$7 - 3 = 4$$ So, $$= \frac{\sqrt{7}\sqrt{7} + \sqrt{7}\sqrt{3}}{4} = \frac{7 + \sqrt{21}}{4}$$ **Answer:** $\frac{7 + \sqrt{21}}{4}$ **(c) $\frac{7}{3 + \sqrt{7}}$** Multiply numerator and denominator by conjugate $3 - \sqrt{7}$: $$\frac{7}{3 + \sqrt{7}} \times \frac{3 - \sqrt{7}}{3 - \sqrt{7}} = \frac{7(3 - \sqrt{7})}{9 - 7}$$ Calculate denominator: $$9 - 7 = 2$$ So, $$= \frac{21 - 7\sqrt{7}}{2}$$ **Answer:** $\frac{21 - 7\sqrt{7}}{2}$