Show that : $\dfrac{1}{(3 - \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})} = 5$

Given,

Equation : $\dfrac{1}{(3 - \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})}$

Simplifying L.H.S. of the above equation :

$\Rightarrow \dfrac{1}{(3 - \sqrt{8})} \times \dfrac{(3 + \sqrt{8})}{(3 + \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})} \times \dfrac{(\sqrt{7} + \sqrt{6})}{(\sqrt{7} + \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} \times \dfrac{(\sqrt{5} + 2)}{(\sqrt{5} + 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} \times \dfrac{(\sqrt{8} + \sqrt{7})}{(\sqrt{8} + \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})} \times \dfrac{(\sqrt{6} + \sqrt{5})}{(\sqrt{6} + \sqrt{5})} \\[1em] \Rightarrow \dfrac{3 + \sqrt{8}}{3^2 - (\sqrt{8})^2} + \dfrac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} + \dfrac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} - \dfrac{\sqrt{8} + \sqrt{7}}{(\sqrt{8})^2 - (\sqrt{7})^2} - \dfrac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \dfrac{3 + \sqrt{8}}{9 - 8} + \dfrac{\sqrt{7} + \sqrt{6}}{7 - 6} + \dfrac{\sqrt{5} + 2}{5 - 4} - \dfrac{\sqrt{8} + \sqrt{7}}{8 - 7} - \dfrac{\sqrt{6} + \sqrt{5}}{6 - 5} \\[1em] \Rightarrow 3 + \sqrt{8} + \sqrt{7} + \sqrt{6} + \sqrt{5} + 2 - (\sqrt{8} + \sqrt{7}) - (\sqrt{6} + \sqrt{5}) \\[1em] \Rightarrow 3 + 2 + \sqrt{8} - \sqrt{8} + \sqrt{7} - \sqrt{7} + \sqrt{6} - \sqrt{6} + \sqrt{5} - \sqrt{5} \\[1em] \Rightarrow 5$

Hence, proved that

$\dfrac{1}{(3 - \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})} = 5$.