<s>[INST] Evaluation of $5\times 5$ determinant The following $5\times 5$ det. comes from a Russian book. I don't want to expand the det. rather than do some operations on it and extract the result.
Prove:
$$\begin{vmatrix}
-1 &1  &1  &1  &x \\ 
 1& -1 &1  &1  &y \\ 
 1& 1 & -1 & 1 &z \\ 
 1& 1 & 1 & -1 & u\\ 
 x& y & z & u &0 
\end{vmatrix}= -4 [x^2+y^2+z^2 +u^2 - 2(xy+zx+zu+yz+yu+\color{red}{x}u)]$$
I have tried many things e.g $R_1 \leftrightarrow R_5+ x R_1$ and of course cyclic ($R_2 \leftrightarrow R_5+ y R_2$ etc) and then adding the last row to the second etc. resulting in:
$$\begin{vmatrix}
0 &y+x  &z+x  &u+x  &x^2 \\ 
 y+x&0  &z+y  &u+y  &y^2 \\ 
 x+z& z+y & 0 &z+u  & z^2\\ 
 u+x& u+y &u+z  & 0 &u^2 \\ 
x &y  &z  &u  &0 
\end{vmatrix}$$
but does not look that promising. 
On the other hand if I add the the first column to the others I do get a lot of zeros, since the first row for example takes the form $-1, 0, 0, 0, x+1$. This looks more promising. But in either case I get stuck. 
May I have some hints or answers? 
Edit: The red letters as suggested by Peter.
 [/INST] I am not sure if this is the most efficient way to do it but doing the following row opperations:
$R_2 \to R_2 + R_1, R_3 \to R_3 + R_1, R_4 \to R_4 + R_1, R_5 \to R_5 + xR_1$
Assuming i didnt make any silly mistakes (as is always the case with there types of problems)
$\begin{vmatrix}
-1&1&1&1&x\\
1&-1&1&1&y\\
1&1&-1&1&z\\
1&1&1&-1&u\\
x&y&z&u&0\\
\end{vmatrix}$ $\to$
$\begin{vmatrix}
-1&1&1&1&x\\
0&0&2&2&x+y\\
0&2&0&2&x+z\\
0&2&2&0&x+u\\
0&x+y&x+z&x+u&x^2\\
\end{vmatrix}$ $\to$
$-\begin{vmatrix}
0&2&2&x+y\\
2&0&2&x+z\\
2&2&0&x+u\\
x+y&x+z&x+u&x^2\\
\end{vmatrix}$
Now removing any one of the $2's$ from any row or column and expanding further into a linear combination of $3\times3$ matrices... Havent finished it but this is the most promising version I've tried
</s>