Anubhav Sinha

If x^2+y^2+z^2=xy+yz+zx then,

2x^2+2y^2+2z^2=2xy+2yz+2zx ie

(x^2+y^2) + (y^2+z^2) + (z^2+x^2) = 2xy+2yz+2zx ie

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

(x-y)^2 + (y-z)^2 + (z-x)^2 = 0 ,

which will happen only when

x = y = z = k (let)

Now, consider

(x+y+z)^3 = x^3 + (y+z)^3 + 3x(y+z) (x+ y +z )

                 =x^3 + (y^3 + z^3 + 3y^2z + 3yz^2 ) +3x (xy + y^2 + yz + xz + yz + z^2)

            = x^3 + y^3 + z^3 + 3y^2z + 3yz^2  +3x^2y + 3xy^2 + 3xyz + 3x^2z + 3xyz +3xz^2

            = x^3 + y^3 + z^3 + 3y^2z + 3yz^2  +3x^2y + 3xy^2 + 3xyz + 3x^2z + 3xyz +3xz^2

so,

x^3 + y^3 + z^3 = (x+y+z)^3 - (3y^2z +3yz^2  +3x^2y +3xy^2 +3xyz +3x^2z +3xyz +3xz^2)

                          = (3k)^3 - k^3 (3+3+3+3+3+3+3+3)

                          = k^3 (27-24)

                          = 3k^3 = 3x^3 = 3y^3 = 3z^3 = 3x^2y= 3x^2z = 3y^2x = 3y^2z = 3z^2x = 3z^2y = 3xyz