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 September 5th, 2013, 11:08 AM #1 Member   Joined: Oct 2011 Posts: 81 Thanks: 0 A rational expression Let x,y,z be three real numbers which satisfy x+y+z=1; xyz=1; x^2+y^2+z^2=2013 Then find the value of 1/(xy+z) + 1/(xz+y) + 1/(yz+x). Please help.
 September 5th, 2013, 11:50 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: A rational expression I would first combine terms in the expression for which we are to determine the value: $\frac{1}{xy+z}+\frac{1}{xz+y}+\frac{1}{yz+x}=\frac {xyz(x+y+z-3)+(x+y+z+1)(xy+xz+yz)}{xyz$$x^2+y^2+z^2+xyz+1$$+( xy)^2+(xz)^2+(yz)^2}$ Using the given values, this becomes: $\frac{1}{xy+z}+\frac{1}{xz+y}+\frac{1}{yz+x}=\frac {-2+2(xy+xz+yz)}{2015+(xy)^2+(xz)^2+(yz)^2}$ Next, take: $x+y+z=1$ Square both sides to get: $x^2+y^2+z^2+2(xy+xz+yz)=1$ Hence: $2(xy+xz+yz)=-2012$ $xy+xz+yz=-1006$ Square again: $(xy)^2+(xz)^2+(yz)^2+2xyz(x+y+z)=1012036$ Hence: $(xy)^2+(xz)^2+(yz)^2=1012034$ And so we have: $\frac{1}{xy+z}+\frac{1}{xz+y}+\frac{1}{yz+x}=\frac {-2014}{2015+1012034}=-\frac{2}{1007}$
 September 5th, 2013, 07:02 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,945 Thanks: 1135 Math Focus: Elementary mathematics and beyond Re: A rational expression I observed that x, y and z are the roots of the equation m³ - m² - 1006m - 1 = 0, but that didn't get me too far.

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### xyz-1=x^2 y^2 z^2

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