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 ?

This is also true in the more general case where $B$ is rank one matrix and can be expressed as $B=uv^T$ for two vectors $u,v\in\mathbb{R}^n$.
When $A$ is invertible, we have that (determinant lemma, https://en.wikipedia.org/wiki/Matrix_determinant_lemma)
$$\det(A+xB)=\det(A+xuv^T)=\det(A)(1+xv^TA^{-1}u)=:\alpha x+\beta,$$
and when $A$ is singular then
$$\det(A+xB)=\det(A+xuv^T)=\det(A)+xv^T\mathrm{Adj}(A)u=:\alpha x+\beta,$$
where $\mathrm{Adj}(A)$ is the adjugate mnatrix of $A$.
