Let $1 = d_1 < d_2 < \dots < d_k = n$ be all positive divisors of the number $n$. We see that
$$d_1 \cdot d_k = d_2 \cdot d_{k-1} = d_3 \cdot d_{k-2} = n.$$
From this we conclude that the product of all positive divisors of $n$ is equal to $n^3$ if and only if $n$ has exactly six divisors.

Positive integer $n$ can be represented in the form $n = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_r^{\alpha_r}$, where $p_1, p_2, \dots, p_r$ are distinct primes, and $\alpha_1 \ge \alpha_2 \ge \dots \ge \alpha_r$ are positive integers.

It is known that the number of the positive divisors of that number is $(\alpha_1 + 1)(\alpha_2 + 1)\dots(\alpha_r + 1)$. Hence, we get
$$(\alpha_1 + 1)(\alpha_2 + 1)\dots(\alpha_r + 1) = 6.$$
Since every factor on the left hand side is greater than 1, there are only two possibilities:
$r = 1, \alpha_1 = 5 \quad \text{or} \quad r = 2, \alpha_1 = 2, \alpha_2 = 1.$

Condition that $n$ has exactly six divisors is satisfied if and only if $n = p^5$ for a prime $p$ or $n = p^2q$ for two distinct primes $p$ and $q$.