The numbers $a$ and $b$ are positive integers and satisfy $96 a^{2}=b^{3}$. What is the smallest value of $a$?
Factoring 96 we have: $2^{5} \times 3 \times a^{2}=b^{3}$. For $2^{5} \times 3 \times a^{2}$ to be a cube, the number $a$ must be of the form: $a=2^{n} \times 3^{m}$. Thus,

$$
2^{5} \times 3 \times a^{2}=2^{5} \times 3 \times\left(2^{n} \times 3^{m}\right)^{2}=2^{5+2 n} \times 3^{1+2 m}=b^{3}
$$

Therefore, $5+2 n$ and $1+2 m$ are multiples of 3. The smallest values of $n$ and $m$ are: $n=2$ and $m=1$. Therefore, $a=2^{2} \times 3=12$.