If x y z 0 then x 2 y 2 z 2 x-y 2 y-z 2 z-x 2 A 0

(E) substitute your own values satisfying x+y+z=0 , let x=0,y=1,z=-1 and solve you get 1/3, substitute any values by your choice you get the same ans . let x=1,y=1,z=-2 . and any other . so none is the answer
10 years agoHelpfull: Yes(14) No(0)
=>(x + y + z)² = x² + y² + z² + 2xy + 2xz + 2yz
=> x² + y² + z² = - 2(xy + xz + yz) .....(1)

(x² + y² + z²)
=>Given, ------------------- = ?
(x-y)²+(y-z)²+(z-x)²

(x² + y² + z²)
=>So, ------------------------------------
2(x² + y² + z²)- 2(xy + xz + yz)

(x² + y² + z²)
=> ----------------------------------- [:: from (1)]
2(x² + y² + z²) + x² + y² + z²

(x² + y² + z²) 1
=> ------------------ = - [::if x,y,z != 0]
3(x² + y² + z²) 3

=> ans (d)None
10 years agoHelpfull: Yes(2) No(2)
e.none
(x+y+z)^2=
10 years agoHelpfull: Yes(0) No(0)
Option (E)
10 years agoHelpfull: Yes(0) No(0)
option E(none)
10 years agoHelpfull: Yes(0) No(0)
given
x+y+z=0
and we know that
(x+y+z)^2=X^2+y^2+z^2+2(xy+yz+zx)
and
from first eqation x+Y+z=0
so
x^2+y^2+z^2=-2(xy+yz+zx)
now to find the value of given eq.
(x-y)^2+(y-z)^2+(z-x)^2=2(x^2+y^2+z^2)-2(xy+yz+zx)
puting the value of eq 2 in above eq we get
2(-2(xy+yz+zx)-2(xy+yx+zx)=-8(xy+yx+zx)
so now putting this value in the que we get
-2(xy+yz+zx)/-8(xy+yz+zx)=1/4
so answer is none E.
10 years agoHelpfull: Yes(0) No(3)
ans is E .. use (a+b+c)^2 and (a-b)^2
10 years agoHelpfull: Yes(0) No(0)

If x + y + z = 0 then,
x^2 + y^2 + z^2/(x-y)^2 + (y-z)^2 + (z-x)^2 = ?

A.0
B.1
C.xyz
D.x+y+z
E.None