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.

$$ \begin{vmatrix}
x & a & b &c \\ 
a & x & b &c \\ 
m &n  & x &p \\ 
 m&  n&  p& x
\end{vmatrix}=0 $$
Subtract the first row to the second row, subtract the third row from the fourth row:
$$ \begin{vmatrix}
x & a & b &c \\ 
a-x & x-a & 0 &0 \\ 
m &n  & x &p \\ 
 0&  0&  p-x& x-p
\end{vmatrix}=0 $$
Factorize $(a-x)$ and $(x-p)$ out:
$$ (a-x)(x-p)\begin{vmatrix}
x & a & b &c \\ 
1 & -1 & 0 &0 \\ 
m &n  & x &p \\ 
 0&  0&  -1& 1
\end{vmatrix}=0 $$
Add the first column to the second column:
$$ (a-x)(x-p)\begin{vmatrix}
x & a+x & b &c \\ 
1 & 0 & 0 &0 \\ 
m &n+m  & x &p \\ 
 0&  0&  -1& 1
\end{vmatrix}=0 $$
Compute the determinant by using the second row:
$$ (a-x)(x-p)\begin{vmatrix}
  a+x & b &c \\  
 n+m  & x &p \\ 
   0&  -1& 1
\end{vmatrix}=0 $$
Add the third column to the second column:
$$ (a-x)(x-p)\begin{vmatrix}
  a+x & b+c &c \\  
 n+m  & p+x &p \\ 
   0&  0& 1
\end{vmatrix}=0 $$
Expand the determinant by the last row:
$$ (a-x)(x-p)\begin{vmatrix}
  a+x & b+c  \\  
 n+m  & p+x  \\ 
\end{vmatrix}=0 $$
Adding the first row to second row:
$$ (a-x)(x-p)\begin{vmatrix}
  a+x & b+c  \\  
 x+a+n+m  & b+c+p+x  \\ 
\end{vmatrix}=0 $$
Factorize $(x+a+n+m)$ out since $a+m+n=b+c+p$:
$$ (a-x)(x-p)(x+a+n+m)\begin{vmatrix}
  a+x & b+c  \\  
1 & 1  \\ 
\end{vmatrix}=0 $$
$$(a-x)(x-p)(x+a+n+m)(x+a-b-c)=0$$
