Find all $a,b,c \in \mathbb{N}$ such that 
\[a^2b|a^3+b^3+c^3,\qquad b^2c|a^3+b^3+c^3, \qquad c^2a|a^3+b^3+c^3.\]

[PS: The original problem was this:
Find all $a,b,c \in \mathbb{N}$ such that 
\[a^2b|a^3+b^3+c^3,\qquad b^2c|a^3+b^3+c^3, \qquad \color{red}{c^2b}|a^3+b^3+c^3.\]
But I think the author meant $c^2a|a^3+b^3+c^3$, just because of symmetry]
1. **Assume gcd(a, b, c) = 1**: 
   We start by assuming that \( \gcd(a, b, c) = 1 \). This assumption simplifies the problem because if \( \gcd(a, b, c) \neq 1 \), then we could divide \( a, b, \) and \( c \) by their greatest common divisor, reducing the problem to the case where \( \gcd(a, b, c) = 1 \).

2. **Symmetry and WLOG \( a \geq b \geq c \)**:
   Without loss of generality, we can assume \( a \geq b \geq c \) due to the symmetry in the problem. This helps in reducing the number of cases we need to consider.

3. **Rewrite the divisibility conditions**:
   Given the conditions:
   \[
   a^2b \mid a^3 + b^3 + c^3, \quad b^2c \mid a^3 + b^3 + c^3, \quad c^2a \mid a^3 + b^3 + c^3,
   \]
   we can combine these into a single equation by noting that \( a^2b^2c^2 \) must divide \( a^3 + b^3 + c^3 \). Let \( d \) be a natural number such that:
   \[
   a^2b^2c^2d = a^3 + b^3 + c^3.
   \]

4. **Inequality analysis**:
   Since \( a \geq b \geq c \), we have:
   \[
   a^2b^2c^2d \leq 3a^3 \implies 3a \geq b^2c^2.
   \]
   This inequality provides a constraint on \( b \) and \( c \) in terms of \( a \).

5. **Further simplification**:
   From the given conditions, we also have:
   \[
   a^2 \mid b^3 + c^3 \implies a^2 \leq b^3 + c^3.
   \]
   Substituting \( b^2c^2 \leq 3a \) into the inequality \( a^2 \leq b^3 + c^3 \), we get:
   \[
   \frac{b^4c^4}{9} \leq b^3 + c^3.
   \]

6. **Case analysis for \( c \)**:
   We analyze the inequality \( \frac{b^4c^4}{9} \leq b^3 + c^3 \) for different values of \( c \):
   - If \( c \geq 2 \), the inequality \( \frac{b^4c^4}{9} \leq b^3 + c^3 \) is generally false for large \( b \). Therefore, \( c \) must be 1.

7. **Case analysis for \( b \)**:
   With \( c = 1 \), the inequality simplifies to:
   \[
   \frac{b^4}{9} \leq b^3 + 1.
   \]
   We test values of \( b \) from 1 to 9:
   - For \( b = 1 \):
     \[
     \frac{1^4}{9} = \frac{1}{9} \leq 1^3 + 1 = 2 \quad \text{(True)}
     \]
   - For \( b = 2 \):
     \[
     \frac{2^4}{9} = \frac{16}{9} \leq 2^3 + 1 = 9 \quad \text{(True)}
     \]
   - For \( b = 3 \):
     \[
     \frac{3^4}{9} = \frac{81}{9} = 9 \leq 3^3 + 1 = 28 \quad \text{(True)}
     \]
   - For \( b = 4 \):
     \[
     \frac{4^4}{9} = \frac{256}{9} \approx 28.44 \leq 4^3 + 1 = 65 \quad \text{(True)}
     \]
   - For \( b = 5 \):
     \[
     \frac{5^4}{9} = \frac{625}{9} \approx 69.44 \leq 5^3 + 1 = 126 \quad \text{(True)}
     \]
   - For \( b = 6 \):
     \[
     \frac{6^4}{9} = \frac{1296}{9} = 144 \leq 6^3 + 1 = 217 \quad \text{(True)}
     \]
   - For \( b = 7 \):
     \[
     \frac{7^4}{9} = \frac{2401}{9} \approx 266.78 \leq 7^3 + 1 = 344 \quad \text{(True)}
     \]
   - For \( b = 8 \):
     \[
     \frac{8^4}{9} = \frac{4096}{9} \approx 455.11 \leq 8^3 + 1 = 513 \quad \text{(True)}
     \]
   - For \( b = 9 \):
     \[
     \frac{9^4}{9} = 6561/9 = 729 \leq 9^3 + 1 = 730 \quad \text{(True)}
     \]

8. **Finding corresponding \( a \)**:
   For each \( b \) from 1 to 9, we need to find \( a \) such that \( a^2 \mid b^3 + 1 \). This is straightforward since \( a \) must be a divisor of \( b^3 + 1 \).

The final answer is \( \boxed{ (a, b, c) = (1, 1, 1) } \) and other solutions can be found by multiplying \( a, b, \) and \( c \) by a constant.