WAEC Further Mathematics 2018 Paper

$x - \;\frac{3}{x^{2}} = x - 3x^{-2}$

Let the power on x be t, so that the power on $x^{-2}$ = 9 - t

$(x)^{t}(x^{-2})^{9 - t} = x^{3} \implies t - 18 + 2t = 3$

$3t = 3 + 18 = 21 \therefore t = 7$

To obtain the coefficient of $x^{3}$, we have

${}^{9}C_{7}(x)^{7}(3x^{-2})^{2} = \;\frac{9!}{(9-7)!\,7!}(x)^{7}(9x^{-4})$

= $\;\frac{9 \times 8 \times 7!}{7!\,2!} \times 9(x^{3}) = 324x^{3}$