find value of determinant $f(n)=β^n+α^n$
then $$
        \begin{vmatrix}
        3 & 1+f(1) & 1+f(2) \\
        1+f(1) & 1+f(2) & 1+f(3) \\
        1+f(2) & 1+f(3) & 1+f(4) \\
        \end{vmatrix}
$$
I don't quite understand which property of determinant can we use here? i tried using linear property but i did not get the required answer.

There is a trick, indeed. The values defined by $f(n)=\alpha^n+\beta^n$ obey the recurrence relation
$$ f(n+2) - (\alpha+\beta)\,f(n+1) + (\alpha\beta)\,f(n) = 0$$
hence the original determinant equals
$$ (1-\alpha)(1-\beta)\det\begin{pmatrix}1+f(0) & 1+f(1) & 1+f(2) \\ 1+f(1) & 1 + f(2) & 1+f(3) \\ 1 & 1 & 1\end{pmatrix}$$
or
$$ (1-\alpha)(1-\beta)\det\begin{pmatrix}f(0) & f(1) & f(2) \\ f(1) & f(2) & f(3) \\ 1 & 1 & 1\end{pmatrix}$$
or
$$ (1-\alpha)^2(1-\beta)^2\det\begin{pmatrix}f(0) & f(1) \\ f(1) & f(2)\end{pmatrix}$$
or
$$ \boxed{\det\begin{pmatrix}1+f(0) & 1+f(1) & 1+f(2) \\ 1+f(1) & 1 + f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4)\end{pmatrix}=\color{red}{(1-\alpha)^2(1-\beta)^2 (\alpha-\beta)^2}.}\tag{1}$$

The properties exploited are


*

*$\det\begin{pmatrix} r_1 \\ r_2 \\ r_3\end{pmatrix}=\det\begin{pmatrix} r_1 \\ r_2 \\ r_3-j r_2+k r_1\end{pmatrix}=\det\begin{pmatrix}r_1-r_3 \\ r_2-r_3 \\ r_3\end{pmatrix}$

*$\det\begin{pmatrix} r_1 \\ r_2 \\ \alpha r_3\end{pmatrix}=\alpha\det\begin{pmatrix} r_1 \\ r_2 \\ r_3\end{pmatrix}$

*$\det M^T = \det M$

*$\det\begin{pmatrix}a & b & 0 \\ c & d & 0 \\ e & f & 1\end{pmatrix}=\det\begin{pmatrix}a & b \\ c & d\end{pmatrix}.$



Another tricky approach may be the following one. One may notice that the determinant is a polynomial in $\mathbb{Z}[\alpha,\beta]$ with degree $4+2=6$, and that the given matrix has rank one if $\alpha=1,\beta=1$ or $\alpha-\beta$. It follows that the determinant is an integer multiple of $(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2$ and by performing an explicit computation at $\alpha=2,\beta=3$ the claim $(1)$ follows.
