<s>[INST] Solve equation in determinant Let $ a,b,c,m,n,p\in \mathbb{R}^{*} $, $ a+m+n=p+b+c $. Solve the equation:
$$ 
\begin{vmatrix}
x & a & b &c \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
 m&  n&  p& x
\end{vmatrix}
=0
$$
I had used the Schur complement ($\det(M)=\det(A)\cdot (D-C\cdot A^{-1}\cdot B)$, for $ M= \begin{bmatrix}
A &B \\ 
C & D
\end{bmatrix}) $ but it didn't help me.
 [/INST] Subtract the first line from the second:
$$\begin{vmatrix}
x & a & b &c \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
 m&  n&  p& x
\end{vmatrix}=\begin{vmatrix}
x-a & a-x & 0 &0 \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
m & n&  p& x
\end{vmatrix}=(x-a)\begin{vmatrix}
1 &-1 & 0 &0 \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
m & n&  p& x
\end{vmatrix}=$$
$1.$ Multiply the first by $-a$ and add on the second
$2.$ Multiply the first by $-m$ and add on the third
$3.$ Multiply the first by $-m$ and add on the forth
$$(x-a)\begin{vmatrix}
1 &-1 & 0 &0 \\ 
0 & x+a & b &c \\ 
0 &m+n  & x &p \\ 
0 &m+ n&  p& x
\end{vmatrix}=$$
Forth line minus third:
$$(x-a)\begin{vmatrix}
1 &-1 & 0 &0 \\ 
0 & x+a & b &c \\ 
0 &m+n  & x &p \\ 
0 &0&  p-x& x-p
\end{vmatrix}=(x-a)(x-p)\begin{vmatrix}
1 &-1 & 0 &0 \\ 
0 & x+a & b &c \\ 
0 &m+n  & x &p \\ 
0 &0&  -1& 1
\end{vmatrix}=$$
Laplace theorem on the first column
$$(x-a)(x-p)\begin{vmatrix}
x+a & b &c \\ 
m+n  & x &p \\ 
0&  -1& 1
\end{vmatrix}=0$$
Add second and third column on the second
$$(x-a)(x-p)\begin{vmatrix}
x+a & b+c &c \\ 
m+n  & x+p &p \\ 
0&  0& 1
\end{vmatrix}=0$$
Laplace on the third line
$$(x-a)(x-p)\begin{vmatrix}
x+a & b+c \\ 
m+n  & x+p \\ 
\end{vmatrix}=0$$
Now $a+m+n=p+b+c=k$ then
$$(x-a)(x-p)\begin{vmatrix}
x+a & k-p \\ 
k-a  & x+p \\ 
\end{vmatrix}=0$$
Add first line on the second
$$(x-a)(x-p)\begin{vmatrix}
x+a & k-p \\ 
x+k  & x+k \\ 
\end{vmatrix}=(x-a)(x-p)(x+k)\begin{vmatrix}
x+a & k-p \\ 
1  & 1 \\ 
\end{vmatrix}=0$$
$$(x-a)(x-p)(x+k)(x+a+p-k)=0$$
$$(x-a)(x-p)(x+k)(x+a+p-k)=0$$
</s>