(x+y+z)^2 = 0 = x^2+y^2+z^2 + 2(xy+xz+yz)
xy+xz+yz = -3
not sure if this helps; try something else
The 2 equations are a plane and a sphere; they intersect at an ellipse.
z = -(x+y)
x^2+y^2+(x+y)^2=6
x^2 + xy + y^2 = 3 contains the information of both equations
Solve quadratic:
x = [-y ± sqrt(y^2 + 12 - 4y^2)]/2
x = [-y ± √3*sqrt(4 - y^2)]/2 {-2 <= x,y <= 2}
not sure if this helps; try something else
The expression: |(x-y)(y-z)(z-x)| substitute z = -(x+y)
= |(x-y)(y+(x+y))(-(x+y)-x)|
= |(x-y)(x+2y)(-1)(2x+y)|
= |2x^3 + 3x^2y - 3xy^2 - 2y^3|
Take partial derivatives of the above:
6x^2 + 6xy - 3y^2 = 0
3x^2 - 6xy - 6y^2 = 0
9(x^2 - y^2) = 0
(x-y)(x+y) = 0
(skipping a few steps; not looking at second partial derivatives)
So either x=y or x= -y
If x = y, the expression is zero; z = -2x
If x = -y, the expression = -2x^3; z = 0
Suppose x = -y and z = 0
x^2+y^2+z^2 = 6 becomes
x^2 + x^2 = 6 --> (x,y) = (√3,-√3) or vice versa
|(x-y)(y-z)(z-x)| becomes 2√3 * √3 * √3 = 6√3
The maximum of |(x-y)(y-z)(z-x)| is 6√3 (maybe)
False starts then ? solution
