I wanna find a closed form of determinant of the following matrix

$$A(n) = \begin{pmatrix} B_{1} & B_{2} & \cdots & B_{n} & 1 \\ B_{n} & B_{1} & \cdots & B_{n-1} &1 \\ \vdots & \vdots & \ddots & \vdots \\ B_{2} & B_{3} & \cdots & B_{1} & 1\\ A_{1} & A_{2} & \cdots & A_{n} &1 \end{pmatrix}$$

For instance :

for $n=3$

$$A(3) = \begin{pmatrix} B_{1} & B_{2} & B_{3} &1 \\ B_{3} & B_{1} & B_{2}&1 \\ B_{2} & B_{3} & B_{1}&1 \\ A_{1} & A_{2} & A_{3} & 1 \end{pmatrix}$$

and

$\det \bigg[A(n)\bigg]= \left(-A_1-A_2-A_3+B_1+B_2+B_3\right)\left(B_1^2-\left(B_2+B_3\right) B_1+B_2^2+B_3^2-B_2 B_3\right) $

I used *Mathematica* for some $n$ and I guessed that determinant of $A(n)$ has a form
$$\color{blue}{\det \bigg[A(n)\bigg]=\bigg(\sum_{i=1}^{n}B_i-\sum_{i=1}^{n}A_i\bigg)\cdot F(B_k)}$$

where $f(B_k)$ is a function of only $B_k$, it doesn't depend on $A_k$.

How to prove this and how to find closed form of $F$ ?

Thank you in advance to any one who may be able to give me some ideas