The numbers $a$ and $b$ are positive integers that satisfy $96 a^{2}=b^{3}$. What is the smallest possible 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 have, at least, the factorization $2^{n} \times 3^{n}$. To find the smallest value of $a$, we take $a=2^{n} \times 3^{n}$, and thus,

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

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