Determinant of an interesting Toeplitz matrix Let $ab=1$. Find
$$\begin{vmatrix} c & a & a^2 & ... & a^{n-1} \\ b & c & a & \dots & a^{n-2} \\ b^2 & b &  c& \dots &a^{n-3}  \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ b^{n-1} & b^{n-2} & b^{n-3} & \dots & c \end{vmatrix}$$
I tried to decompose by line however it does not lead to a solution

Let $M$ be the matrix at hand. Its entries have the form
$$M_{ij} = (c-1)\delta_{ij} + b^{i-j} = (c-1)\delta_{ij} + b^i a^j$$
where $\delta_{ij} = \begin{cases}1, & i = j\\ 0,& i \ne j\end{cases}$ is the Kronecker delta. We can express $M$ as a rank-1 update of the diagonal matrix $(c-1)I_n$:
$$M = (c-1)I_n + uv^T
\quad\text{ where }\quad
\begin{cases}
u &= (b,b^2,\ldots,b^n)^T\\
v &= (a,a^2,\ldots,a^n)^T
\end{cases}
$$
When $c \ne 1$, by matrix determinant lemma, we have:
$$\det(M) = (c-1)^n \det\left(I_n + \frac{1}{c-1}uv^T\right)
= (c-1)^n\left( 1 + \frac{1}{c-1}v^Tu\right)$$
Since $v^T u = \sum\limits_{k=1}^n a^k b^k = \sum\limits_{k=1}^n 1  = n$, we obtain
$$\det(M) = (c-1)^n\left( 1 + \frac{n}{c-1}\right) =  (c-1)^{n-1}(c + n - 1)\tag{*1}$$
When $c = 1$, $M = uv^T$ implies the columns of $M$ are linearly dependent. This forces $\det(M) = 0$ and formula $(*1)$ works at $c = 1$.
