WAEC Further Mathematics 2018 Paper

Let the power of $2x^{2}$ be t and the power of $\;\frac{1}{x} \equiv x^{-1}$ = 9 - t.

$(2x^{2})^{t}(x^{-1})^{9 - t} = x^{0}$

Dealing with x alone, we have

$(x^{2t})(x^{-9 + t}) = x^{0} \implies 2t - 9 + t = 0$

$3t - 9 = 0 \therefore t = 3$

The binomial expansion is then,

${}^{9}C_{3} (2x^{2})^{3}(x^{-1})^{6} = \;\frac{9!}{(9-3)! \; 3!} \times 2^{3}$

= 84 x 8

= 672