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.

Let $Q$ be a real orthogonal matrix whose first column is $(\frac12,\frac12,\frac12,\frac12)^T$. Let also $v=(x,y,z,u)^T$ and $Q^Tv=(a,b,c,d)$. Then the matrix in question is equal to $\pmatrix{Q&0\\ 0&1}A\pmatrix{Q^T&0\\ 0&1}$, where
$$
A=\left[\begin{array}{cccc|c}
2&&&&a\\ &-2&&&b\\ &&-2&&c\\ &&&-2&d\\
\hline
a&b&c&d&0
\end{array}\right].
$$
Hence the required determinant is equal to $\det(A)$. Let $e_1=(1,0,0,0)^T$ and $E\in M_4(\mathbb R)$ be the all-one matrix. By Laplace expansion along the last column of $A$, we see that
$$
\det(A)=8(a^2-b^2-c^2-d^2).
$$
Now we can rewrite this in terms of $x,y,z$ and $u$:
\begin{align}
\det(A)&=8(a^2-b^2-c^2-d^2)\\
&=16a^2-8(a^2+b^2+c^2+d^2)\\
&=16(v^TQe_1)(e_1Q^Tv)-8(v^TQ)(Q^Tv)\\
&=16v^T\left(\frac14E\right)v-8v^Tv\\
&=4(x+y+z+u)^2 - 8(x^2+y^2+z^2+u^2).
\end{align}
