How to solver this equation $\sum_{i=1}^{6}x_{i}x_j=-3,j=1,2,3,4,5,6,j\neq i$ Let  $x_{i}\in R,i=1,2,3,4,5,6$ such that 
$$\begin{cases}
x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{1}x_{5}+x_{1}x_{6}=-3\\
x_{2}x_{1}+x_{2}x_{3}+x_{2}x_{4}+x_{2}x_{5}+x_{2}x_{6}=-3\\
x_{3}x_{1}+x_{3}x_{2}+x_{3}x_{4}+x_{3}x_{5}+x_{3}x_{6}=-3\\
x_{4}x_{1}+x_{4}x_{2}+x_{4}x_{3}+x_{4}x_{5}+x_{4}x_{6}=-3\\
x_{5}x_{1}+x_{5}x_{2}+x_{5}x_{3}+x_{5}x_{4}+x_{5}x_{6}=-3\\
x_{6}x_{1}+x_{6}x_{2}+x_{6}x_{3}+x_{6}x_{4}+x_{6}x_{5}=-3
\end{cases}$$
Find the $x_{i}$.
My idea: let
$$S=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}$$
then
$$x_{i}(S-x_{i})=-3,i=1,2,3,4,5,6$$
then follow I fell ugly,Thank you

Following on David Peterson's hint:  This shows there are at most two different values among the $x_i$ (solve the quadratic).  Call them $a,b$.  We go through the cases:  
1)all the same.  Each equation becomes $5a^2=-3$, which gives $x_i=\pm\sqrt{\frac 35}i$, which is acceptable over $\Bbb C$ but not $\Bbb R$  
2)five the same, one different.  Let the odd one be $a$.  We then get $5ab=-3, b(a+4b)=ab+4b^2-3, 4b^2=4ab$, one of $a,b$ is zero, which fails  
3)four the same, two the other.  Let the four be $a$.  We get $b(4a+b)=4ab+b^2=-3=a(3a+2b)=3a^2+2ab, (b+3a)(b-a)=0$  We reject $a=b$ as we covered that in case 1. $-3a^2=-3$ so $a=\pm1, b=\mp3$  
4)three of one, three of the other.  Summing $x_i=\frac 12(S \pm \sqrt{12+S^2})$ and noting three of the signs are plus and three minus gives $S=3S, S=0$ and $a=-b$.  This gives $-a^2=-3$ and we have three $x$'s are $\sqrt 3$ and three are $-\sqrt 3$
