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

1.	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» (d) 0 Given, x + y + z = 0 ⇒ x + y = – z ⇒ x²+ y² + 2xy = z² ⇒ x² + y² = z² – 2xy ∴ \(\frac{1}{x^2+y^2-z^2} = \frac{1}{z^2-2xy-z^2}=\frac{1}{-2xy}=-\frac{1}{2xy}\) Similarly, \(\frac{1}{y^2+z^2-x^2} = -\frac{1}{2xy}\) and \(\frac{1}{z^2+x^2-y^2}=-\frac{1}{2zx}\) ∴ \(\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}\) = \(-\frac{1}{2xy}-\frac{1}{2yz}-\frac{1}{2zx}\) = \(-\frac{1}{2}\big[\frac{z+x+y}{xyz}\big]\) = 0. [∵ x + y + z = 0]

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»

(d) 0

Given, x + y + z = 0 ⇒ x + y = – z

⇒ x²+ y² + 2xy = z² ⇒ x² + y² = z² – 2xy

∴ \(\frac{1}{x^2+y^2-z^2} = \frac{1}{z^2-2xy-z^2}=\frac{1}{-2xy}=-\frac{1}{2xy}\)

Similarly, \(\frac{1}{y^2+z^2-x^2} = -\frac{1}{2xy}\) and \(\frac{1}{z^2+x^2-y^2}=-\frac{1}{2zx}\)

∴ \(\frac{1}{x^2+y^2-z^2}+\frac{1}{y^2+z^2-x^2}+\frac{1}{z^2+x^2-y^2}\)

= \(-\frac{1}{2xy}-\frac{1}{2yz}-\frac{1}{2zx}\) = \(-\frac{1}{2}\big[\frac{z+x+y}{xyz}\big]\) = 0. [∵ x + y + z = 0]