Is this problem based on discrete polynomials modulo $(x^3-1)$ strong? We start working with the Ring $R=\left(\mathbb{Z}/p\mathbb{Z}\right)\left[x\right]/\left(x^{3}-1\right)$,
$p$ prime, i.e. degree two polynomials with coefficients modulo $p$
modulo $x^{3}-1$. As $x^{3}-1=\left(x-1\right)\left(x^{2}+x+1\right)$,
we choose a subset of $R$, $S\subset R$, and a public value, $z\in\mathbb{Z}/p\mathbb{Z},\ z\neq0$,
$P\in S,\ P\equiv-z\left(mod\ x-1\right)$. 

We can define the transpose operation of elements of $S$ as swaping
the $x$ coefficient with the $x^{2}$one of the corresponding polynomial,
so $\left(ax^{2}+bx+c\right)^{T}=bx^{2}+ax+c$.

Now, we define a function $f:S\times S\mapsto S$, as $f\left(A,B\right)=\left(xA^{T}+z\right)\left(xB^{T}+z\right)-zx$.
It's null element is $-zx$ and $f$ is a closed map of $S$, so $A,B\in S,\ f\left(A,B\right)\in S$. 

Next, we define a series as follows:

$A,B\in S,\ s_{0}=A,\ s_{1}=B,\ s_{n}=f\left(s_{n-2},s_{n-1}\right)$

And for a given element of the series,$s_{n}$, a value $r_{n}=f\left(s_{n},A\right)$

The question is

Taking into account that the function $f$ is not
associative, how difficult is, knowing $B$ and $r_{n}$, recover the
value of a secret $A$. As an example of sizes let's say $n=256,\ p\sim2^{128}$.

This problem can lead to a cryptosystem described in this two documents:

https://drive.google.com/open?id=1OGnFfooWASVCD1Iw_hVwvHYgqMGGE5nH

https://drive.google.com/open?id=1OeKh_ZJF-i7_KzWFRv8jodk3YkXe2qyv
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.