If xy+yz+zx=0 then find the value of 1/x^2-yz + 1/y^2-zx + 1/z^2-xy
Malyarpita Singh , 6 Years ago
Grade 10
10 grade maths> If xy+yz+zx=0 then find the value of 1/x^...
1 Answers
Arun
Last Activity: 6 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
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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