Question

x^2+y^2+z^2-xy-yz-zx=0 and find x+y/z

Answer 1

Given, x^2+y^2+z^2-xy-yz-zx=0 can be written as

        =   2(x^2+y^2+z^2-xy-yz-zx) = 2 * 0

        =   2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx = 0

        = x^2 + x^2 + y^2 + y^2 + z^2 + z^2 - 2xy - 2yz - 2zx = 0

       = x^2 + y^2 - 2xy + y^2 + z^2 - 2yz + z^2 + x^2 - 2xz = 0

We know that (a-b)^2 = a^2 + b^2 - 2ab

      = (x-y)^2+(y-z)^2+(z-x)^2=0

      = x - y=0, y - z=0, z - x=0

      = x = y, y = z, z = x

     Let x = y = z = a
         
     So, (x+y)/z = (a+a)/a

                       = 2a/a

                       = 2.


Hope this helps!