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If $$x-\frac{1}{x}=y, \qquad y-\frac{1}{y}=z, \qquad z-\frac{1}{z}=x$$ find the value of $xyz$.

This is how far I proceeded: $x+y+z=z-1/z+x-1/x+y-1/y=>1/x+1/y+1/z=0 =>xy+yz+zx=0$
Also from question, $x^2-1=xy,y^2-1=yz,z^2-1=zx$.
Adding $x^2+y^2+z^2-3=xy+yz+zx=0 =>x^2+y^2+z^2=3$ .
I am stuck here please help. This image gives some hint, but I am unable to understand it.

Shubham Kumar Verma
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    Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. – 5xum Aug 01 '19 at 08:28
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    Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. Even if the question is closed, you can still edit it, and we will vote to reopen it. – 5xum Aug 01 '19 at 08:28
  • I have updated my approach to question. – Shubham Kumar Verma Aug 01 '19 at 09:13
  • Which part of the solution do you not understand? – Toby Mak Aug 01 '19 at 09:33
  • Where it says 'add all'. How did we get the zero first? – Shubham Kumar Verma Aug 01 '19 at 09:34
  • You seemed to have entered the wrong equation. Signs are negative; you entered positive. – Shubham Kumar Verma Aug 01 '19 at 09:41
  • Oh, that explains why. – Toby Mak Aug 01 '19 at 09:41

4 Answers4

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Given: $x-\frac1x=y,y-\frac1y=z,z-\frac1z=x$, you found correctly: $$x-\frac1x+y-\frac1y+z-\frac1z=y+z+x \Rightarrow \frac1x+\frac1y+\frac1z=0 \Rightarrow xy+yz+zx=0 \ \ (1)\\ x^2-1+y^2-1+z^2-1=xy+yz+zx \Rightarrow x^2+y^2+z^2=3 \ \ (2)$$ Square each and add them all: $$x^2+\frac1{x^2}-2+y^2+\frac1{y^2}-2+z^2+\frac1{z^2}-2=y^2+z^2+x^2 \Rightarrow \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=6 \Rightarrow \\ x^2y^2+y^2z^2+z^2x^2=6x^2y^2z^2 \ \ (3)$$ Square $(1)$: $$\underbrace{x^2y^2+y^2z^2+z^2x^2}_{6x^2y^2z^2}+2xyz(x+y+z)=0 \Rightarrow x+y+z=-3xyz \ \ (4)$$ Square $(4)$: $$\underbrace{x^2+y^2+z^2}_{3}+2(\underbrace{xy+yz+zx}_{0})=9x^2y^2z^2 \Rightarrow 3=9x^2y^2z^2 \Rightarrow xyz=\pm \frac1{\sqrt{3}}.$$

farruhota
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Hint: By the equation #(3) we get in #(1) $$z-1/z-\frac{1}{z-\frac{1}{z}}=y$$ and with $z=y-\frac{1}{y}$ we get $$3y^6-9y^4+6y^2-1=0$$ Multiplying all terms together gives $$x^2(1-z^2)+y^2(1-z^2)+z^2(1-y^2)=0$$ substituting $$xy=x^2-1$$ $$yz=y^2-1$$ $$xz=z^2-1$$ so we get $$xyz(x+y+z)=0$$

Dr. Sonnhard Graubner
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Note that adding up the three equations gives you:

$$x+y+z =x+y+z - \frac{1}{x} - \frac{1}{y} - \frac{1}{z}$$

, so $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$$

Try it from there

David
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Eliminating $x$ and $z$ between the equations we obtain: $$3y^6-9y^4+6y^2-1=0.$$

By symmetry we know that $x$ and $z$ must also satisfy this equation. Hence $x^2$, $y^2$ and $z^2$ are roots of $$3y^3-9y^2+6y-1=0.$$

Therefore $(xyz)^2=\frac{1}{3}$ and hence $xyz=\pm\frac{1}{\sqrt 3}$.

In addition I think this may be one of the simplest direct solutions:

Adding the equations: $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=>xy+yz+zx=0.$$ Clearing the denominators of the original equations ($x^2-xy-1=0, \dots$) and adding we obtain $$x^2+y^2+z^2=xy+yz+zx+3=3$$ Hence $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=3=>(x+y+z)=\pm\sqrt{3}$$Rearranging the original equations ($y+\frac{1}{x}=x,\dots$), multiplying them and simplifying: $$(y+\frac{1}{x})(z+\frac{1}{y})(x+\frac{1}{z})=xyz =>(x+y+z)=-\frac{1}{xyz}.$$ Combining we obtain the result $$xyz=\pm\frac{1}{\sqrt{3}}.$$

Ivan
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  • You can only deduce $(xyz)^2=\frac13$ if $x^2,y^2,z^2$ are distinct roots of the cubic. This is not difficult to show, from the original equations, but it has to be done for a complete proof. – TonyK Aug 01 '19 at 10:15
  • But how can I proceed through solution using the method I provided in the question(image link). I am unable to understand the steps after 'add all'. how are we getting zero after that? – Shubham Kumar Verma Aug 01 '19 at 10:21
  • @ShubhamKumarVerma: I would say that is not Ivan's problem! – TonyK Aug 01 '19 at 10:27
  • @TonyK Subtle, but fair point! In this case as you say it's trivial to see that x=y is not possible, similarly for the other equalities. – Ivan Aug 01 '19 at 11:23
  • @Ivan: Not quite so simple. You must also rule out $y=-x$ etc. – TonyK Aug 01 '19 at 11:36