Find all positive integers $n$ such that $n^3$ is the product of all divisors of $n$.
To solve the problem, we need to find all positive integers \( n \) such that \( n^3 \) is the product of all divisors of \( n \).

1. **Understanding the Product of Divisors**:
   Let \( n \) be a positive integer and let \( d_1, d_2, \ldots, d_{\tau(n)} \) be the divisors of \( n \), where \( \tau(n) \) is the number of divisors of \( n \). The product of all divisors of \( n \) can be written as:
   \[
   \prod_{d \mid n} d
   \]
   Notice that for each divisor \( d \) of \( n \), \( \frac{n}{d} \) is also a divisor of \( n \). Therefore, we can pair each divisor \( d \) with \( \frac{n}{d} \), and the product of each pair is \( n \):
   \[
   d \cdot \frac{n}{d} = n
   \]
   Since there are \( \tau(n) \) divisors, and each pair multiplies to \( n \), the product of all divisors is:
   \[
   \prod_{d \mid n} d = n^{\tau(n)/2}
   \]

2. **Given Condition**:
   According to the problem, \( n^3 \) is the product of all divisors of \( n \):
   \[
   n^3 = n^{\tau(n)/2}
   \]

3. **Equating Exponents**:
   By equating the exponents, we get:
   \[
   3 = \frac{\tau(n)}{2}
   \]
   Solving for \( \tau(n) \):
   \[
   \tau(n) = 6
   \]

4. **Finding \( n \) with \( \tau(n) = 6 \)**:
   We need to find all positive integers \( n \) such that \( \tau(n) = 6 \). The number of divisors function \( \tau(n) \) for \( n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \) is given by:
   \[
   \tau(n) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]
   We need to find combinations of \( e_1, e_2, \ldots, e_k \) such that:
   \[
   (e_1 + 1)(e_2 + 1) \cdots (e_k + 1) = 6
   \]
   The possible factorizations of 6 are:
   - \( 6 = 6 \times 1 \)
   - \( 6 = 3 \times 2 \)
   - \( 6 = 2 \times 3 \)

   These correspond to:
   - \( n = p^5 \) (where \( p \) is a prime)
   - \( n = p_1^2 p_2 \) (where \( p_1 \) and \( p_2 \) are distinct primes)

5. **Special Case**:
   We should also consider the special case \( n = 1 \). The number 1 has exactly one divisor (itself), and the product of its divisors is \( 1 \). Thus, \( 1^3 = 1 \), which satisfies the condition.

Therefore, the positive integers \( n \) that satisfy the given condition are:
\[
n = 1, \quad n = p^5, \quad n = p_1^2 p_2
\]
where \( p, p_1, \) and \( p_2 \) are distinct primes.

The final answer is \( \boxed{ n = 1, \quad n = p^5, \quad n = p_1^2 p_2 } \).