Question

If $x+y+z=0$ then what is the value of
$\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}$

• A. $\frac{1}{x^2+y^2+z^2}$
• B. $1$
• C. $-1$
• D. $0$

Answer 1

The correct option is D $0$

$x+y+z=0\Rightarrow x+y=-z$

On squaring, we get

$\to (x+y{)}^2=(-z{)}^2$

$\to x^2+y^2-z^2=-2xy$

Similarly,

$y^2+z^2-x^2=-2yz$

$z^2+x^2-y^2=-2xz$

$\therefore \frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}$

= $\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}=\frac{-1}{2}\left(\frac{z+x+y}{xyz}\right)=0$