<s>[INST] Determinant of a matrix with $t$ in all off-diagonal entries. It seems from playing around with small values of $n$ that
$$
\det \left( \begin{array}{ccccc}
-1 & t & t & \dots & t\\
t & -1 & t & \dots & t\\
t & t & -1 & \dots & t\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t  & t & t & \dots& -1
\end{array}\right) = (-1)^{n-1}(t+1)^{n-1}((n-1)t-1)
$$
where $n$ is the size of the matrix.
How would one approach deriving (or at least proving) this formally?

Motivation
This came up when someone asked what is the general solution to:
$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b},$$
and for non-trivial solutions, the matrix above (with $n=3$) must be singular. In this case either $t=-1\implies a+b+c=1$ or $t=\frac{1}{2}\implies a=b=c$.
So I wanted to ensure that these are also the only solutions for the case with more variables.
 [/INST] Using elementary operations instead of induction is key.
$$\begin{align}
&\begin{vmatrix}
-1 & t  & t  & \dots & t\\
t  & -1 & t  & \dots & t\\
t  & t  & -1 & \dots & t\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t  & t & t & \dots& -1
\end{vmatrix}\\
&=
\begin{vmatrix}
-t-1 & 0 & 0 & \dots & t+1\\
0 & -t-1 & 0 & \dots & t+1\\
0 & 0 & -t-1 & \dots & t+1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t  & t & t & \dots& -1
\end{vmatrix}\\
&=
\begin{vmatrix}
-t-1 & 0 & 0 & \dots & 0\\
0 & -t-1 & 0 & \dots & 0\\
0 & 0 & -t-1 & \dots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t  & t & t & \dots& (n - 1)t -1
\end{vmatrix}\\
&=
(-1)^{n - 1}(t + 1)^{n - 1}((n - 1)t - 1)
\end{align}$$
</s>