A) 0 |
B) 1 |
C) -1 |
D) \( \Large \frac{1}{2} \) |
A) 0 |
Given : \( \Large x+y+z=0 \)
=> \( \Large x+y = -z \)
Squaring \( \Large \left(x+y\right)^{2}=z^{2} \)
=> \( \Large x^{2}+y^{2}+2xy=z^{2} \)
=> \( \Large x^{2}+y^{2}-z^{2} = -2xy \)
Similarly \( \Large y^{2}+z^{2}-x^{2} = -2yz \)
and \( \Large z^{2}+x^{2}-y^{2} = -2zx \)
Therefore, \( \Large \frac{1}{x^{2}+y^{2}-z^{2}}+\frac{1}{y^{2}+z^{2}-x^{2}}+\frac{1}{z^{2}+x^{2}-y^{2}} \)
= \( \Large - \frac{1}{2xy} - \frac{1}{2yz} - \frac{1}{2zx} \)
= \( \Large - \left(\frac{x+y+z}{2xyz}\right) \)
\( \Large =0 \) ...\( \Large \because x+y+z=0 \)