Question 102

If $$x,y,z \neq 0$$ and $$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}$$ = $$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$$ then the relation among x, y, z is

Solution

Expression : $$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}$$ = $$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}$$

Taking L.C.M on both sides

= $$\frac{x^2 + y^2 + z^2}{x^2 y^2 z^2} = \frac{xy + yz + zx}{x^2 y^2 z^2}$$

= $$x^2 + y^2 + z^2 - xy - yz - zx = 0$$

= $$\frac{1}{2} [(x-y)^2 + (y-z)^2 + (z-x)^2] = 0$$

=> $$(x = y)$$ and $$(y = z)$$ and $$(z = x)$$

=> $$x = y = z$$

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