Determine whether it is possible to partition $\mathbb{Z}$ into triples $(a,b,c)$ such that, for every triple, $|a^3b + b^3c + c^3a|$ is perfect square.
1. **Given Condition**: We need to determine if it is possible to partition the set of integers $\mathbb{Z}$ into triples $(a, b, c)$ such that for every triple, $|a^3b + b^3c + c^3a|$ is a perfect square.

2. **Expression Simplification**: Consider the condition $a + b + c = 0$. We can simplify the expression $a^3b + b^3c + c^3a$ as follows:
   \[
   a^3b + b^3c + c^3a = a^3b + b^3(-a - b) + (-a - b)^3a
   \]
   Expanding $(-a - b)^3$:
   \[
   (-a - b)^3 = -a^3 - 3a^2b - 3ab^2 - b^3
   \]
   Substituting this back into the expression:
   \[
   a^3b + b^3(-a - b) + (-a - b)^3a = a^3b + b^3(-a - b) + (-a^3 - 3a^2b - 3ab^2 - b^3)a
   \]
   Simplifying further:
   \[
   = a^3b - a^3b - b^4 - 3a^2b^2 - 3ab^3 - b^3a - a^4 - 3a^3b - 3a^2b^2 - 2ab^3 - b^4
   \]
   \[
   = -a^4 - 2a^3b - 3a^2b^2 - 2ab^3 - b^4
   \]
   \[
   = -(a^2 + ab + b^2)^2
   \]
   Therefore, $a^3b + b^3c + c^3a = -(a^2 + ab + b^2)^2$, which is the negative of a perfect square.

3. **Partitioning $\mathbb{Z}$**: We need to show that it is possible to partition $\mathbb{Z}$ into triples $(a, b, c)$ such that $a + b + c = 0$.

4. **Example Partition**: Consider the set $\{-1, 0, 1\}$. Clearly, $-1 + 0 + 1 = 0$.

5. **General Partitioning Strategy**: We need to partition the set $A = \{2, 3, \dots\}$ into triples of the form $\{a, b, a+b\}$. We can do this in an infinite number of steps:
   - At each step, remove $u$, $v$, and $u+v$, where $u$ and $v$ are the smallest unused elements of $A$.
   - This ensures that each element of $A$ is placed in exactly one triple.

6. **Verification**: For any triple $(a, b, a+b)$, we have:
   \[
   a + b + (a+b) = 2a + 2b = 0 \quad \text{(mod 2)}
   \]
   This ensures that the sum of the elements in each triple is zero.

Conclusion:
\[
\boxed{\text{Yes, it is possible.}}
\]