If positive integers, $a,b,c$ are pair-wise co-prime, and, \[\ a^2|(b^3+c^3), b^2|(a^3+c^3), c^2|(a^3+b^3) \] find $a,b,$ and $c$
1. Given that \(a, b, c\) are pairwise coprime positive integers, we have the following conditions:
   \[
   a^2 \mid (b^3 + c^3), \quad b^2 \mid (a^3 + c^3), \quad c^2 \mid (a^3 + b^3)
   \]

2. Since \(a, b, c\) are pairwise coprime, the product \(a^2 b^2 c^2\) must divide the sum \(a^3 + b^3 + c^3\). This is because \(a^2, b^2, c^2\) are coprime integers dividing \(a^3 + b^3 + c^3\).

3. Let's analyze the divisibility conditions one by one:
   - \(a^2 \mid (b^3 + c^3)\)
   - \(b^2 \mid (a^3 + c^3)\)
   - \(c^2 \mid (a^3 + b^3)\)

4. Consider the first condition \(a^2 \mid (b^3 + c^3)\). Since \(a\) is coprime with \(b\) and \(c\), \(a\) must divide \(b^3 + c^3\). Similarly, for the other conditions, we have:
   - \(b^2 \mid (a^3 + c^3)\) implies \(b \mid (a^3 + c^3)\)
   - \(c^2 \mid (a^3 + b^3)\) implies \(c \mid (a^3 + b^3)\)

5. Since \(a, b, c\) are pairwise coprime, the only way for \(a^2, b^2, c^2\) to divide \(a^3 + b^3 + c^3\) is if \(a, b, c\) are all equal to 1. This is because any higher value would not satisfy the coprime condition and the divisibility simultaneously.

6. Therefore, the only solution that satisfies all the given conditions is:
   \[
   a = 1, \quad b = 1, \quad c = 1
   \]

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