Explanation

(b)
$In\ the\ expansion\ of\ \left( x^{2} + \frac{a}{x} \right)^{5},\ the\ general\ term\ is$
$T_{r + 1} = \ ^{5}C_{r}(x^{2})^{5 - r}\left( \frac{a}{x} \right)^{r} = \ ^{5}C_{r} \bullet a^{r} \bullet x^{10 - 3r}$
For the coefficient of $x,$ put
$\ \ \ \ \ \ \ \ \ \ \ \ \ 10 - 3r = 1\ \Longrightarrow r = 3$
$\therefore\$Coefficient of $x = \ ^{5}C_{3}a^{3} = {10a}^{3}$