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Question Number 74912 by behi83417@gmail.com last updated on 03/Dec/19

$$\{\begin{array}{l}{\mathbf{x}}^2=\mathbf{yz}+1\\ {\mathbf{y}}^2=\mathbf{xz}+2\\ {\mathbf{z}}^2=\mathbf{xy}+3\end{array}\Rightarrow \mathbf{x}+\mathbf{y}+\mathbf{z}=?$$

Answered by MJS last updated on 03/Dec/19

$$y=px\wedge z=qx$$$$\left(1\right)(1-pq)x^2-1=0\Rightarrow x^2=\frac{1}{1-pq}$$$$\left(2\right)(p^2-q)x^2-2=0\Rightarrow x^2=\frac{2}{p^2-q}$$$$\left(3\right)(q^2-p)x^2-3=0\Rightarrow x^2=\frac{3}{q^2-p}$$$$\frac{1}{1-pq}=\frac{2}{p^2-q}\Rightarrow q=-\frac{p^2-2}{2p-1}$$$$\left(1\right),\left(2\right)x^2=\frac{2p-1}{p^3-1}$$$$\left(3\right)x^2=\frac{3{(2p-1)}^2}{(p^3-1)(p-4)}$$$$\frac{2p-1}{p^3-1}=\frac{3{(2p-1)}^2}{(p^3-1)(p-4)}\Rightarrow p=-\frac{1}{5}\vee p=\frac{1}{2}$$$$p=\frac{1}{2}\Rightarrow q\mathrm{not}\mathrm{defined}$$$$p=-\frac{1}{5}\Rightarrow q=-\frac{7}{5}$$$$\Rightarrow x=\pm \frac{5\sqrt{2}}{6}\wedge y=\mp \frac{\sqrt{2}}{6}\wedge z=\mp \frac{7\sqrt{2}}{6}$$$$x+y+z=\pm \frac{\sqrt{2}}{2}$$

Commented by behi83417@gmail.com last updated on 04/Dec/19

$$\mathrm{dear}\mathrm{proph}:\mathrm{MJS}!\mathrm{it}\mathrm{is}\mathrm{great}.\mathrm{thank}\mathrm{you}$$$$\mathrm{very}\mathrm{much}\mathrm{sir}.$$

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