WAEC Further Mathematics 2016 Paper

Given $\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}}$,

first, we rationalise by multiplying through with $2\sqrt{3} - 3\sqrt{2}$ (the inverse of the denominator).

$\left(\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}}\right)\left(\frac{2\sqrt{3} - 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}}\right)$

= $\frac{16\sqrt{3} - 24\sqrt{2} - 18\sqrt{2} + 18\sqrt{3}}{4(3) - 6\sqrt{6} + 6\sqrt{6} - 9(2)}$

= $\frac{34\sqrt{3} - 42\sqrt{2}}{-6} = 7\sqrt{2} - \frac{17\sqrt{3}}{3}$