Find all triples $(a,b,c)$ of positive integers such that
(i) $a \leq b \leq c$;
(ii) $\text{gcd}(a,b,c)=1$; and
(iii) $a^3+b^3+c^3$ is divisible by each of the numbers $a^2b, b^2c, c^2a$.
To solve the problem, we need to find all triples \((a, b, c)\) of positive integers that satisfy the given conditions:
1. \(a \leq b \leq c\)
2. \(\text{gcd}(a, b, c) = 1\)
3. \(a^3 + b^3 + c^3\) is divisible by each of \(a^2b\), \(b^2c\), and \(c^2a\)

Let's analyze each condition step-by-step.

1. **Condition 3: Divisibility by \(a^2b\)**

   Since \(a^3 + b^3 + c^3\) is divisible by \(a^2b\), we have:
   \[
   a^2b \mid a^3 + b^3 + c^3
   \]
   This implies:
   \[
   a^2b \mid b^3 + c^3 \quad \text{(since } a^3 \text{ is clearly divisible by } a^2b \text{)}
   \]
   Therefore, \(a^2b \mid b^3 + c^3\).

2. **Condition 3: Divisibility by \(b^2c\)**

   Similarly, since \(a^3 + b^3 + c^3\) is divisible by \(b^2c\), we have:
   \[
   b^2c \mid a^3 + b^3 + c^3
   \]
   This implies:
   \[
   b^2c \mid a^3 + c^3 \quad \text{(since } b^3 \text{ is clearly divisible by } b^2c \text{)}
   \]
   Therefore, \(b^2c \mid a^3 + c^3\).

3. **Condition 3: Divisibility by \(c^2a\)**

   Similarly, since \(a^3 + b^3 + c^3\) is divisible by \(c^2a\), we have:
   \[
   c^2a \mid a^3 + b^3 + c^3
   \]
   This implies:
   \[
   c^2a \mid a^3 + b^3 \quad \text{(since } c^3 \text{ is clearly divisible by } c^2a \text{)}
   \]
   Therefore, \(c^2a \mid a^3 + b^3\).

4. **Analyzing the implications**

   From the above conditions, we can infer that:
   \[
   a^2b \mid b^3 + c^3
   \]
   \[
   b^2c \mid a^3 + c^3
   \]
   \[
   c^2a \mid a^3 + b^3
   \]

   Since \(a, b, c\) are pairwise coprime and must all be odd (as shown in the initial solution), let's consider the simplest case where \(a = 1\).

5. **Case \(a = 1\)**

   If \(a = 1\), then the conditions simplify to:
   \[
   1^2b \mid b^3 + c^3 \implies b \mid b^3 + c^3
   \]
   \[
   b^2c \mid 1^3 + c^3 \implies b^2c \mid 1 + c^3
   \]
   \[
   c^2 \cdot 1 \mid 1^3 + b^3 \implies c^2 \mid 1 + b^3
   \]

   Since \(b \mid b^3 + c^3\), this is always true. Now, we need to check the other two conditions.

6. **Checking \(b^2c \mid 1 + c^3\)**

   For \(b^2c \mid 1 + c^3\), since \(b\) and \(c\) are coprime, \(b\) must divide \(1 + c^3\). This implies \(b = 1\) (since \(b\) is a positive integer and must divide 1).

7. **Checking \(c^2 \mid 1 + b^3\)**

   For \(c^2 \mid 1 + b^3\), since \(b = 1\), this simplifies to \(c^2 \mid 2\). The only positive integer \(c\) that satisfies this is \(c = 1\).

Therefore, the only solution is \((a, b, c) = (1, 1, 1)\).

The final answer is \( \boxed{ (1, 1, 1) } \).