Question 255011
Given, x+y+z=1 and x^2+y^2+z^2=3
Now, (x+y+z)^2 = x^2+y^2+z^2+2(xy+xz+yz)
         1^2 = 3+2(xy+yz+zx)
        1-3  = 2(xy+yz+zx)
          -2 = 2(xy+yz+zx)
          -2/2 = xy+yz+zx
     xy+yz+zx = -1