Is there a way to simplify $\det(D + C)$, where $D,C$ are square matrices of matching dimensions, $D$ is diagonal (with different diagonal elements, $D_{ij} = \delta_{ij}d_i$), and $C$ is a constant matrix, that is, all entries $C_{ij}=c$ are equal to the same number?
To be more explicit, assuming $D,C\in\mathbb{R}^{n\times n}$, the matrix $D+C$ has the form:
$$D + C = \left(\begin{array}{ccccc} d_1 + c & c & c & \cdots & c\\ c & d_2 + c & c & \cdots & c\\ c & c & d_3 + c & \cdots & c\\ \vdots & \vdots & \vdots & & \vdots\\ c & c & c & \cdots & d_n + c \end{array}\right)$$
Subtracting the first row from all other rows (determinant does not change under this operation):
$$ \left(\begin{array}{cccccc} d_1 + c & c & c & \cdots & c & c\\ - d_1 & d_2 & 0 & 0 & \cdots & 0\\ - d_1 & 0 & d_3 & 0 & \cdots & 0\\ - d_1 & 0 & 0 & d_4 & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & & \vdots\\ - d_1 & 0 & 0 & 0 & \cdots & d_n \end{array}\right) $$
The determinant in this form is readily evaluated by expanding in minors. The result is
$$\det(C+D)=\left( 1 + c \sum_{i = 1}^n d_i^{- 1} \right) d_1 d_2 \ldots d_n$$
Thanks @coffeemath & @daw for the hint!