Let $\alpha$ and $\beta $ be fixed non-zero reals and $f(n)=\alpha^n+\beta^n$ with $$A=\begin{pmatrix} 3&1+f(1)&1+f(2)\\1+f(1)&1+f(2)&1+f(3)\\ 1+f(2)&1+f(3)&1+f(4) \end{pmatrix}.$$ How to find determinant of the matrix $A$ i.e. $|A|,$ in the simplest form? Firstly I thought that matrix may be of a particular form. But its neither circulant matrix nor Vandermonde type or tridiagonal type. Then I tried multilinear property on columns of given matrix as below:
$$\det\begin{pmatrix} 3&1+f(1)&1+f(2)\\1+f(1)&1+f(2)&1+f(3)\\ 1+f(2)&1+f(3)&1+f(4) \end{pmatrix}=\det\begin{pmatrix} 3&1+f(1)&1\\1+f(1)&1+f(2)&1\\ 1+f(2)&1+f(3)&1 \end{pmatrix} +\det\begin{pmatrix} 3&1+f(1)&f(2)\\1+f(1)&1+f(2)&f(3)\\ 1+f(2)&1+f(3)&f(4) \end{pmatrix} $$ I further apply multilinear property on columns but didn't get my answer. Please help. Thanks.
Hint: Consider $$ \left(\begin{matrix} 1&1&1\\1&\alpha&\beta\\ 1^2&\alpha^2&\beta^2 \end{matrix}\right)\left(\begin{matrix} 1&1&1^2\\1&\alpha&\alpha^2\\ 1&\beta&\beta^2 \end{matrix}\right). $$ Do you recognize it now?
Possibly more systematic way: Recognizing each column of $A$ is a linear combination of vectors$$ u=\left(\begin{matrix}1\\ 1 \\1 \end{matrix}\right),\quad \ v=\left(\begin{matrix}1\\\alpha\\ \alpha^2\end{matrix}\right),\quad \ w=\left(\begin{matrix}1\\ \beta\\ \beta^2\end{matrix}\right),$$ express the matrix as $$ A=(u+\ v+ w\ ,\ \ u+\alpha v+\beta w\ ,\ \ u+\alpha^2 v+ \beta^2 w). $$ Regarding $u,v,w$ as if they were numbers, if necessary, $$ A=(u,v,w)\left(\begin{matrix} 1&1&1\\1&\alpha&\alpha^2\\ 1&\beta&\beta^2 \end{matrix}\right)=\left(\begin{matrix} 1&1&1\\1&\alpha&\beta\\ 1&\alpha^2&\beta^2 \end{matrix}\right)\left(\begin{matrix} 1&1&1\\1&\alpha&\alpha^2\\ 1&\beta&\beta^2 \end{matrix}\right). $$