Explanation

(b)
The general term in the expansion of $\left( x + \frac{1}{x^{2}} \right)^{n - 3}$ is given by
$T_{r + 1} = \ ^{n - 3}C_{r}(x)^{n - 3 - r}\left( \frac{1}{x^{2}} \right)^{r}$
$= \ ^{n - 3}C_{r}x^{n - 3 - 3r}$
As $x^{2k}$ occurs in the expansion of $\left( x + \frac{1}{x^{2}} \right)^{n - 3}$, we must have $n - 3 - 3r = 2k$ for some non-negative integer $r$
$\Rightarrow \ 3(1 + r) = n - 2k$
$\Rightarrow \ n - 2k$ is a multiple of 3