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Question Number 166830 by HongKing last updated on 28/Feb/22

Answered by nurtani last updated on 01/Mar/22

(x+(1/y))(y+(1/z))(z+(1/x))=(xy+(x/z)+1+(1/(yz)))(z+(1/x))=xyz+x+z+(1/y)+y+(1/z)+(1/x)+(1/(xyz))= (4)(1)((7/3)) ⇔ xyz+(1/(xyz))+(x+(1/y))+(y+(1/z))+(z+(1/x))=(4)(1)((7/3))=((28)/3) ⇔ xyz+(1/(xyz))+4+1+(7/3)=((28)/3) ⇔ xyz+(1/(xyz))+((22)/3)=((28)/3)⇔ xyz+(1/(xyz))=(6/3)=2 ⇔ x^2 y^2 z^2 +1=2xyz ⇔ (xyz)^2 −2xyz+1=0 let: ϕ=xyz ⇒ ϕ^2 −2ϕ+1=0 ⇒ (ϕ−1)^2 =0 ⇒ ϕ=1 ∴ ϕ = xyz = 1

$$\left({x}+\frac{\mathrm{1}}{{y}}\right)\left({y}+\frac{\mathrm{1}}{{z}}\right)\left({z}+\frac{\mathrm{1}}{{x}}\right)=\left({xy}+\frac{{x}}{{z}}+\mathrm{1}+\frac{\mathrm{1}}{{yz}}\right)\left({z}+\frac{\mathrm{1}}{{x}}\right)={xyz}+{x}+{z}+\frac{\mathrm{1}}{{y}}+{y}+\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{xyz}}=\:\left(\mathrm{4}\right)\left(\mathrm{1}\right)\left(\frac{\mathrm{7}}{\mathrm{3}}\right) \\ $$$$\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}+\left({x}+\frac{\mathrm{1}}{{y}}\right)+\left({y}+\frac{\mathrm{1}}{{z}}\right)+\left({z}+\frac{\mathrm{1}}{{x}}\right)=\left(\mathrm{4}\right)\left(\mathrm{1}\right)\left(\frac{\mathrm{7}}{\mathrm{3}}\right)=\frac{\mathrm{28}}{\mathrm{3}} \\ $$$$\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}+\mathrm{4}+\mathrm{1}+\frac{\mathrm{7}}{\mathrm{3}}=\frac{\mathrm{28}}{\mathrm{3}} \\ $$$$\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}+\frac{\mathrm{22}}{\mathrm{3}}=\frac{\mathrm{28}}{\mathrm{3}}\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}=\frac{\mathrm{6}}{\mathrm{3}}=\mathrm{2} \\ $$$$\Leftrightarrow\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} +\mathrm{1}=\mathrm{2}{xyz} \\ $$$$\Leftrightarrow\:\left({xyz}\right)^{\mathrm{2}} −\mathrm{2}{xyz}+\mathrm{1}=\mathrm{0} \\ $$$${let}:\:\varphi={xyz}\:\Rightarrow\:\varphi^{\mathrm{2}} −\mathrm{2}\varphi+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\left(\varphi−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:\varphi=\mathrm{1} \\ $$$$\:\:\:\:\:\therefore\:\varphi\:=\:{xyz}\:=\:\mathrm{1} \\ $$

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