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in Polynomials by (24.0k points)
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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

1 Answer

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Best 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]

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