Is this problem based on discrete polynomials modulo $(x^3-1)$ strong?
Here's the isomorphism:

$ax^2 + bx + c$ (with $z$) maps to $ax^2 + (b + z - z')x + c$ (with $z'$)

The only nonobvious thing about this transform is that it preserves $f$, that is, that $f(A, B)$ (with $z$) is the element that maps to $f'(A', B')$ (with $z'$, and $f', A', B'$ are the mapped versions of $f, A, B$)

However, this is not hard to show; we notice that the constraints $ax^2 + bx + c = -z \pmod {x-1}$ is equivalent to $a + b + c = -z$.

With this in mind, if $A = ax^2 - (a+b+z)x + b$ (setting the linear term to the value it must be for it to be consistent with the constraint), and if $B = cx^2 - (c+d+z)x + d$, then $f(A, B) = (bd -2ac – ad – bc)x^2 - (-ac+ad+bc+ad+z)x + (2ad+2bc+ac+bd)$

If we consider the mapped version of those two elements $A' = ax^2 - (a+b+z')x + b$ and $B' = cx^2 - (c+d+z')x + d$, then the mapped version of $f'$ would have $f'(A, B) = (bd -2ac – ad – bc)x^2 - (-ac+ad+bc+ad+z')x + (2ad+2bc+ac+bd)$

We can see that the element $f(A, B)$ maps to the element $f'(A', B')$, hence the isomorphism is preserved.