I would like to compute $\operatorname{Tor}_k^{R}(\mathbb{Z},\mathbb{Z})$, where $R = \mathbb{Z}[x]/(x^n -1)=\mathbb Z[X]$.
I think I am near to do so, but I can not figure out the last step. I started looking for a free resolution of $\mathbb{Z}$ as an $R$-module, and found that
$$\cdots\longrightarrow R \longrightarrow R \longrightarrow \mathbb{Z} \longrightarrow 0$$
where $d_0: R\longrightarrow \mathbb{Z}: f(X) \mapsto f(1)$, and for $i\geqslant 1$, $d_{2i-1}: R\longrightarrow R$ is multiplication by $X-1$ and $d_{2i}: R\longrightarrow R$ by $1+X+\cdots+X^{n-1}$
In this way I have a free resolution of $\mathbb{Z}$ as an $R$-module. Now I can apply the functor $ * \otimes \mathbb{Z}$ to the free resolution (projective too):
$$C^*:\cdots \longrightarrow R \otimes \mathbb{Z} \longrightarrow R \otimes \mathbb{Z} \longrightarrow 0$$
where the functions defined above do the same in the first component and apply identity in the second component. Now I know that there is an isomorphism between $R \otimes N$ and $N$ if $N$ is an $R$-module. Then I would obtain something like
$$\cdots \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow 0$$
which it may looks easier to find the $n$-th cohomology object of the cochain complex. But I do not know how to find $\ker(d_n\otimes 1)$ neither $\operatorname{im}(d_n\otimes 1)$.
Note that the isomorphism $R\otimes_R \mathbb Z\to \mathbb Z $ is induced from the left action, so that it sends $r\otimes n\to rn$. If $r$ is a polynomial, then the action is given by evaluation at $1$, that is, $rn = r(1)n$. Under this map, multiplication by $x-1$ goes to the zero map $0:\mathbb Z\to\mathbb Z$, while multiplication by $1+\cdots+x^{n-1}$ goes to multiplication by $n:\mathbb Z\to\mathbb Z$. Can you continue?