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If xy+yz+zx=0 then find the value of 1/x^2-yz + 1/y^2-zx + 1/z^2-xy

If xy+yz+zx=0 then find the value of 1/x^2-yz + 1/y^2-zx + 1/z^2-xy

Grade:10

1 Answers

Arun
25750 Points
5 years ago
Dear student
 
 
xy + yz + zx = 0 

so yz = -x(y+z) 
x^2-yz = x(x+y+z) 

so 1/(x^2-yz) = 1/(x(x+y+z)) 
similarlly 
1/(y^2-xz) = 1/(y(x+y+z)) 
and 
1/(x^2-yz) = 1/(x(x+y+z)) 
1/(z^2-xy) = 1/(x(x+y+z)) 

adding all 3 we get your expression 

1/(x^2 - yz) + 1/(y^2 - zx) +1/(z^2 - xy) = 1/(x+y+z)/(1x+ 1/y + 1/z) 

1/x + 1/y + 1/z = (yz+xz+xy)/xyz = 0 

so 
1/(x^2 - yz) + 1/(y^2 - zx) +1/(z^2 - xy) = 0
 
Regards
Arun (askIITians forum expert)

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