Question

If xy + yz + zx = 0, then the value of
(1/x²-yz +1/y²-zx +1/z²-xy) is

(1) 3
(3) 1
is
x² - yz y² – zx 2² - xy
(2) o
(4) x + y + 2
(Jharkhand, Stage-1, 2016-17)​

Answer 1

xy+yz+zx=0∴xy+zx=−yz⇒xy+yz=−zx⇒yz+zx=−xy∴1x2−yz+1y2−zx+1z2−xy

Putting values of -yz, -zx, -xy from above

⇒1x2+(xy+zx)+1y2+(xy+yz)

      +1z2+(yz+zx)

⇒1x(x+y+z)+1y(x+y+z)

      +1z(x+y+z)

⇒1(x+y+z)(1x+1y+1z)⇒1(x+y+z)(zy+xz+xyxyz)⇒1x+y+z×0⇒0