WAEC Further Mathematics 2016 Paper

Concept:The expression can be written as a binomial $\left(a + b\right)^6$, where the coefficient of $x^{3}$ is found using the general term.Explanation:Rewrite the given expression.$\left[\frac{1}{3}(2 + x)\right]^{6} = \left(\frac{2}{3} + \frac{x}{3}\right)^{6}$The general term for $\left(a + b\right)^{n}$ is ${}^{6}C_{r} a^{6-r} b^{r}$.Here, $a = \frac{2}{3}$ and $b = \frac{x}{3}$.For the coefficient of $x^{3}$, set $r = 3$.So the term is:${}^{6}C_{3}\left(\frac{2}{3}\right)^{3}\left(\frac{x}{3}\right)^{3}$Compute the binomial coefficient:${}^{6}C_{3} = \frac{6!}{3!3!} = 20$Compute the powers:$\left(\frac{2}{3}\right)^{3} = \frac{8}{27}$ and $\left(\frac{x}{3}\right)^{3} = \frac{x^{3}}{27}$Multiply all parts:$20 \times \frac{8}{27} \times \frac{1}{27} = \frac{160}{729}$Thus, the coefficient of $x^{3}$ is $\frac{160}{729}$.Answer:$\frac{160}{729}$ (Option D).