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$x, y, z$ are positive real numbers which satisfy the following three equations.
$$x + 2y + z = 5(x + y)(y + z)$$ $$x + y + 2z = 7(y + z)(z + x)$$ $$2x + y + z = 6(z + x)(x + y)$$

Find the value of $(24)^3(xyz)$.

Okay , so here I'm probably not supposed to solve these equations , but rather find some "trick" to calculate $xyz$. First thing that came to my mind was adding all three equations , which gives :
$$6 x^2+18 x y+18 x z+5 y^2+18 y z+7 z^2 = 4 (x+y+ z)$$
Now what? Any hints are apreciated. (This is not class-homework , I'm solving sample questions for a competitive exam )

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Observe that $$x+y+y+z=x+2y+z $$ set $$x+y=a,y+z=b,z+x=c$$ – lab bhattacharjee Mar 30 '14 at 11:02
@labbhattacharjee I'll try that , thanks. – A Googler Mar 30 '14 at 11:03
@labbhattacharjee I solved the question , thanks a lot for the hint! – A Googler Mar 30 '14 at 11:14
If you have solved the question, please post a solution (and accept it, when the software allows it). It gives you practice in writing things up, and it helps keep the Unanswered Questions list in order. – Gerry Myerson Mar 30 '14 at 11:28
@GerryMyerson I posted the answer . – A Googler Mar 30 '14 at 11:51

Let $ x+y=a,y+z=b,z+x=c$. (From labbhattacharjee's hint in comments)
Hence the equations become ,
$$a + b =5ab$$ $$c+b=7cb$$ $$a+c=6ca$$
These can be easily solved to give ,
$b=1/3, c=1/4,a=1/2$
Now back substitute $a=x+y,b=y+z,c=z+x$
Then the equations become , $$y+z=1/3$$ $$x+y=1/2$$ $$z+x=1/4$$ Now these equation can be easily solved to give , $$z=1/24$$ $$x=5/24$$ $$y=7/24$$ Hence $24*24*24*xyz=35$

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Congratulations and thanks for your post answering your question and showing your effort. Cheers. – Claude Leibovici Mar 30 '14 at 12:50

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