To solve the equation $a^3 + b^3 + c^3 = 2001$ in positive integers $(a, b, c)$, we can explore possible values for $a$, $b$, and $c$ that satisfy this equation.

First, calculate some bounds for the integers. Assume $a \leq b \leq c$, which implies $3a^3 \le 2001$. Thus, we find:
$$a^3 \le \frac{2001}{3} \approx 667 \quad \Rightarrow \quad a \le \sqrt[3]{667} \approx 8.7.$$
Since $a$ must be an integer, we have $a \le 8$.

Next, try possible values for $a$ and find corresponding pairs $(b, c)$ such that the sum of cubes equals 2001.

### Case $a = 1$:
$$b^3 + c^3 = 2000$$
Explore values of $b$:
1. $b = 10$, then $c^3 = 2000 - 10^3 = 2000 - 1000 = 1000 \Rightarrow c = 10$.

Hence, one solution is $(a, b, c) = (1, 10, 10)$.

### Case $a = 10$:
$$b^3 + c^3 = 2001 - 10^3 = 2001 - 1000 = 1001$$
Explore values of $b$:
1. $b = 10$, then $c^3 = 1001 - 10^3 = 1001 - 1000 = 1 \Rightarrow c = 1$.

Thus, another solution is $(a, b, c) = (10, 10, 1)$.

### Symmetric Cases:
The nature of the problem allows permutation of $a$, $b$, and $c$, so we get additional solutions by permuting these variables:
- $(a, b, c) = (10, 1, 10)$.

Therefore, the solutions in positive integers are:
$$\boxed{(a, b, c) = (10, 10, 1), (10, 1, 10), (1, 10, 10)}$$