Let $d(n)$ denote the number of positive divisors of the positive integer $n$. Solve the equation

$$
d\left(n^{3}\right)=n
$$
Solution. For $n=1$, it is clearly a solution, since $d\left(1^{3}\right)=d(1)=1$.

For $n \neq 1$, let the prime factorization of $n$ be $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$, where each $\alpha_{i}>0$ and $p_{1}<p_{2}<\ldots<p_{k}$. We need to show that:

$$
n = d\left(n^{3}\right) = \prod_{i=1}^{k} \left(3 \alpha_{i} + 1\right)
$$

We start by considering the inequality:

$$
1 \leq \frac{3 \alpha_{j} + 1}{p_{j}^{\alpha_{j}}}
$$

For $p_{j} \neq 2$, we have:

$$
\begin{aligned}
1 & \leq \frac{3 \alpha + 1}{5^{\alpha+1}} \\
5(3 \alpha + 1) & > 3 \alpha + 4 \\
12 \alpha + 1 & > 0
\end{aligned}
$$

This shows that for $p_{j} \neq 2$, the inequality $1 \leq \frac{3 \alpha_{j} + 1}{p_{j}^{\alpha_{j}}}$ cannot hold. Therefore, the prime factorization of $n$ must include the prime number 2, with the exponent $\alpha_{1}$ satisfying:

$$
1 \leq \frac{3 \alpha_{1} + 1}{2^{\alpha_{1}}}
$$

This inequality holds for $\alpha_{1}=1, 2, 3$, but not for $\alpha_{1}=4$. Since the sequence $\frac{3 \alpha + 1}{2^{\alpha}}$ is strictly decreasing for positive $\alpha$ (as shown by the above sequence $\frac{3 \alpha + 1}{5^{\alpha}}$), it does not hold for $\alpha_{1} > 4$. Therefore, we have:

$$
n = 2 \cdot p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}} \text{ or } n = 2^{2} \cdot p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}} \text{ or } n = 2^{3} \cdot p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}
$$

a) For $n = 2 \cdot p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$:

$$
n = d\left(n^{3}\right) = 4 \cdot (3 \alpha_{2} + 1) \ldots (3 \alpha_{k} + 1)
$$

This implies that $n$ is divisible by 4. However, this case is not possible because the prime factorization of $n$ includes 2 only to the first power.

b) For $n = 2^{2} \cdot p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$:

$$
n = d\left(n^{3}\right) = 7 \cdot (3 \alpha_{2} + 1) \ldots (3 \alpha_{k} + 1)
$$

This implies that the prime factorization of $n$ includes the prime number 7. If 7 appears to the first power and $n$ has no other prime factors, then $n = 2^{2} \cdot 7 = 28$, and $d\left(n^{3}\right) = (3 \cdot 2 + 1)(3 \cdot 1 + 1) = 28$, which is a solution. If 7 appears to a power greater than 1 or if $n$ has other prime factors, the equality (1) cannot hold because the left-hand side expression would definitely decrease, making it less than 1 (if 7 appears to a higher power, because the sequence $p_{i}(\alpha) = \frac{3 \alpha + 1}{p_{i}^{\alpha}}$ is strictly decreasing, and if $n$ has other prime factors, because $\frac{3 \alpha_{i} + 1}{p_{i}^{\alpha_{i}}} < 1$ if $p_{i} \neq 2$). Therefore, in this case, $n = 28$ is the only solution.

c) For $n = 2^{3} \cdot p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$:

$$
n = d\left(n^{3}\right) = 10 \cdot (3 \alpha_{2} + 1) \ldots (3 \alpha_{k} + 1)
$$

This implies that the prime factorization of $n$ includes the prime number 5. The solution $n = 2^{3} \cdot 5 = 40$ satisfies the condition, and there are no other solutions in this case (for the same reason as in the previous part, the left-hand side expression would decrease, making it less than 1).

Thus, we have shown that the solutions to the problem are $n = 1$, $n = 28$, and $n = 40$.