Question

if x+y+z=0, prove that x^2-yz= y^2-zx=z^2-xz

Answer 1

Given that,
x+y+z=0
x=-(y+z)
squaring on both sides we get,
x^2={-(y+z)}^2
x^2=y^2+z^2+2yz................................1

x+y+z=0
y=-(x+z)
squaring on both sides we get,
y^2=x^2+z^2+2xz...............................2

x+y+z=0
z=-(x+y)
squaring on both sides we get
z^2=x^2+y^2+2xy...............................3

by substituting eq1 in the below equation we get,
x^2-yz=y^2+z^2+2yz-yz
x^2=y^2+z^2+2yz
x^2={-(y+z)}^2
x=-(y+z)
x+y+z=0.................................................4

by substituting eq2 in below equation we get,
y^2-zx=x^2+z^2+2xz-zx
y^2={-(x+z)}^2
y=-(x+z)
x+y+z=0..................................................5

by substituting eq3 in below equation we get,
z^2-xy=x^2+y^2+2xy-xy
z^2=x^2+y^2+2xy
z^2={-(x+y)}
z=-(x+y)
x+y+z=0..................................................6

After solving equations (x^2-yz),(y^2-zx),
(z^2-xy) we are achieving same result as we can see in eq4,5,6 i.e x+y+z=0
In mathematics and science there are no co-incidence everything occurs due to some particular reason.
Hence with no doubt we can conclude that,
x^2-yz=y^2-xz=z^2-xy




Here the equation written in the question is z^2-xz but not z^2-xy
I think it should be z^2-xy,even in proof it can be clearly observed ,also when we observe other equations the squared term is not repeated in the next term like y^2-zx.