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If $x+y+z=6$ and $xyz=2$, then find the value of $$\cfrac{1}{xy} +\cfrac{1}{yz}+\cfrac{1}{zx}$$

I've started by simply looking for a form which involves the given known quantities ,so:

$$\cfrac{1}{xy} +\cfrac{1}{yz} +\cfrac{1}{zx}=\cfrac{yz\cdot zx +xy \cdot zx +xy \cdot yz}{(xyz)^2}$$

Now this might look nice since I know the value of the denominator but if I continue to work on the numerator I get looped :

$$\cfrac{yz\cdot zx +xy \cdot zx +xy \cdot yz}{(xyz)^2}=\cfrac{4\left(\cfrac{1}{xy}+\cfrac{1}{zy}+\cfrac{1}{zy}\right)}{(xyz)^2}=\cfrac{4\left(\cfrac{(\cdots)}{(xyz)^2}\right)}{(xyz)^2}$$

How do I deal with such continuous fraction ?

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  • $\begingroup$ Is $(xyz)^2$ the lowest common denominator? $\endgroup$ – Thomas Andrews Dec 17 '15 at 13:36
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    $\begingroup$ $\frac{1}{xy} +\frac{1}{yz}+\frac{1}{zx}=\frac{x+y+z}{xyz}$. $\endgroup$ – user236182 Dec 17 '15 at 13:37
  • $\begingroup$ lol I've messed up so badly.Shame on me.... $\endgroup$ – Mr. Y Dec 17 '15 at 13:37
  • $\begingroup$ Do I delete the question or leave it (In case someone will do the same mistake (I know that's impossible)) ? $\endgroup$ – Mr. Y Dec 17 '15 at 13:41
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    $\begingroup$ Don't automatically think you should delete a question/answer if it is flawed. People can still learn from it. That's the beauty of making mistakes. $\endgroup$ – MPW Dec 17 '15 at 13:46
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$$\cfrac{1}{xy} +\cfrac{1}{yz} +\cfrac{1}{zx}=\cfrac{yz\cdot zx +xy \cdot zx +xy \cdot yz}{(xyz)^2}=\frac{x+y+z}{xyz}$$

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  • $\begingroup$ I don't know if that's a joke. $\endgroup$ – Mr. Y Dec 17 '15 at 13:39
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    $\begingroup$ don't worry sometime it happens. $\endgroup$ – Kushal Bhuyan Dec 17 '15 at 13:40
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Notice, the method is straight forward simply take L.C.M. & substitute known values as follows $$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{x+y+z}{xyz}=\frac{6}{2}=\mathbb{\color{red}{3}}$$

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$$x+y+z=6$$ Divide both sides with $xyz$ $$\frac{x+y+z}{xyz}=\frac{6}{xyz}$$ $$\frac{x}{xyz}+\frac{y}{xyz}+\frac{z}{xyz}=\frac{6}{2}$$ $$\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}=3$$

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  • $\begingroup$ Is your third equation a typo? $\endgroup$ – YoTengoUnLCD Dec 17 '15 at 13:56

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