How can I prove that $f(x)=\det (A+xB)= \alpha x+\beta $? I have two matrices $A$ and $B$, such that :
$$A=A(a,b,c)=\begin{pmatrix} a & c & c & \dots & c \\ b & a & c & \dots & c\\ b & b & a & \dots & c\\ \vdots &\vdots &\vdots & \ddots &\vdots\\ b & b & b &\dots& a \end{pmatrix} \hspace{1cm} \text{and} \hspace{1cm} B=A(1, 1, 1)$$
And we've :
$$f(x)=\det(A+xB)$$
I have to prove the existence of two real numbers  $\alpha$ and $\beta$, such that :
$$f(x)=\alpha x +\beta$$
Just prove their existence not their values, because later in the same exercise we have to calculate $f(-c)$ and $f(-b)$, then deduce their values, then deduce $\det(A)$.
So That's why I think calculating the determinant won't be a good idea I guess. Any Ideas to do so ?

In $A+xB$, every entry is of the form $const+x$. We do not change thet determinant if we subtract the first column from every other column. After that, only the first columnt depends on $x$, i.e., the first column is of the form $v+xw$ and all other columns are constant. As $\det$ is linear in every column, the result follows.
