Yes, it is possible. If $a + b + c = 0$, then \[a^3b + b^3c + c^3a = a^3b + b^3(-a - b) + (-a - b)^3 a= -a^4 - 2a^3b - 3a^2b^2 - 2ab^3 - b^4 = -(a^2 + ab + b^2)^2\]is the negative of a square. So we show that it is possible to partition $\mathbb{Z}$ into triples summing to 0. First take $\{-1, 0, 1\}$; then it is sufficient to split $A \stackrel{\text{def}}{=} \{2, 3, \dots\}$ into triples of the form $\{a, b, a+b\}$. We do so in an infinite number of steps: at each step we remove $u$, $v$, and $u+v$, where $u$ and $v$ are the smallest unused elements of $A$. It is clear that this places each element of $A$ in exactly one triple.

Proposed by Nikolai Beluhov.