यदि `xy+yz+zx=0`, तो `(1/(x^(2)-yz)+1/(y^(2)-zx)+1/(z^(2)-xy)) (x,y,z!= 0)` किसके बराबर है?

Options

A. 3
B. 1
C. `x+y+z`
D. `0`

Answer

D

Explanation

`xy+yz+zx=0`
`:. xy+zx=-yz`
`implies xy+yz=-zx`
`implies yz+zx=-xy`
`:. 1/(x^(2)-yz)+1/(y^(2)-zx)+1/(z^(2)-xy)`
Putting values of `-yz, -zx, -xy` from above
`implies 1/(x^(2)+(xy+zx))+1/(y^(2)+(xy+yx))`
`+1/(z^(2)+(yx+zx))`
`implies 1/(x(x+y+z))+1/(y(x+y+z)`
`+1/(z(x+y+z))`
`implies 1/((x+y+z))(1/x+1/y+1/z)`
`implies 1/(x+z+z)((zy+xz+xy)/(xyz))`
`implies 1/(x+y+z)xx0=0`