If xy+yz+zx=0, then ((1/(x^2 −yz))+(1/(y^2 −zx))+(1/(z^2 −xy)))(x,y,z ≠ 0) is equal to


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Question Number 112802 by Aina Samuel Temidayo last updated on 09/Sep/20

$If xy + yz + zx = 0, then$ $(\frac{1}{x^{2} - yz} + \frac{1}{y^{2} - zx} + \frac{1}{z^{2} - xy}) (x, y, z \neq 0) is$ $equal to$
	Answered by MJS_new last updated on 09/Sep/20

	$z = - \frac{x y}{x + y} \Rightarrow value = 0$
	Answered by Dwaipayan Shikari last updated on 09/Sep/20

	$\frac{1}{x^{2} + x y + z x} + \frac{1}{y^{2} + y x + x y} + \frac{1}{z^{2} + x z + z y} (x y + z x = - y z . . .)$ $= \frac{1}{(x + y + z)} (\frac{x y + y z + z x}{x y z}) = 0$
	Answered by nimnim last updated on 09/Sep/20
	$xy+yz+zx= { ((−xy=yz+zx)),((−yz=xy+zx)),((−zx=xy+yz)) :} (1/(x^2 −yz))+(1/(y^2 −zx))+(1/(z^2 −xy)) =(1/(x^2 +xy+zx))+(1/(y^2 +xy+yz))+(1/(z^2 +yz+zx)) =(1/(x(x+y+z)))+(1/(y(y+x+z)))+(1/(z(z+y+x))) =((yz+zx+xy)/(xyz(x+y+z)))=(0/(xyz(x+y+z)))=0★$
	$xy + yz + zx = {\begin{cases} - xy = yz + zx \\ - yz = xy + zx \\ - zx = xy + yz \end{cases}$ $\frac{1}{x^{2} - yz} + \frac{1}{y^{2} - zx} + \frac{1}{z^{2} - xy}$ $= \frac{1}{x^{2} + xy + zx} + \frac{1}{y^{2} + xy + yz} + \frac{1}{z^{2} + yz + zx}$ $= \frac{1}{x (x + y + z)} + \frac{1}{y (y + x + z)} + \frac{1}{z (z + y + x)}$ $= \frac{yz + zx + xy}{xyz (x + y + z)} = \frac{0}{xyz (x + y + z)} = 0 ★$
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