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Question Number 93138 by i jagooll last updated on 11/May/20

$$\{\begin{array}{l}{18\mathrm{x}}^2=3\mathrm{y}(1+{9\mathrm{x}}^2)\\ {18\mathrm{y}}^2=3\mathrm{z}(1+{9\mathrm{y}}^2)\\ {18\mathrm{z}}^2=3\mathrm{x}(1+{9\mathrm{z}}^2)\end{array}$$

Answered by john santu last updated on 11/May/20

$$\{\begin{array}{l}{\mathrm{x}}^2=\frac{\mathrm{y}}{6-9\mathrm{y}}\\ {\mathrm{y}}^2=\frac{\mathrm{z}}{6-9\mathrm{z}}\\ {\mathrm{z}}^2=\frac{\mathrm{x}}{6-9\mathrm{x}}\end{array}$$$${\left(\mathrm{xyz}\right)}^2=\frac{\mathrm{xyz}}{(6-9\mathrm{x})(6-9\mathrm{y})(6-9\mathrm{z})}$$$$(6-9\mathrm{x})(6-9\mathrm{y})(6-9\mathrm{z})=\frac{1}{\mathrm{xyz}}$$$$\{\begin{array}{l}6-9\mathrm{x}=\frac{1}{\mathrm{x}}\Rightarrow \mathrm{x}=\frac{1}{3}\\ 6-9\mathrm{y}=\frac{1}{\mathrm{y}}\Rightarrow \mathrm{y}=\frac{1}{3}\\ 6-9\mathrm{z}=\frac{1}{\mathrm{z}}\Rightarrow \mathrm{z}=\frac{1}{3}\end{array}$$$$$$

Answered by MJS last updated on 11/May/20

$$\mathrm{if}x=y=z=\mathrm{c}\left(\mathrm{because}\mathrm{of}\mathrm{symmetry}\right)\Rightarrow$$$$c=0\vee c=\frac{1}{3}$$$$\mathrm{if}x\ne y\ne z\mathrm{we}\mathrm{get}$$$$x\approx -.137591\pm .0972771\mathrm{i}$$$$y\approx .0812039\mp .129974\mathrm{i}$$$$z\approx -.0372460\pm .147390\mathrm{i}$$$$\mathrm{and}\mathrm{these}\mathrm{values}\mathrm{are}\mathrm{interchangeable}$$

Commented by i jagooll last updated on 11/May/20

$$\mathrm{how}\mathrm{get}\mathrm{x}\in \mathbb{C}\mathrm{sir}?$$

Commented by MJS last updated on 11/May/20

$$18x^2=3y(9x^2+1)\Rightarrow y=\frac{6x^2}{9x^2+1}$$$$18y^2=3z(9y^2+1)\Rightarrow z=\frac{6y^2}{9y^2+1}=\frac{216x^4}{405x^4+18x^2+1}$$$$18z^2=3x(9z^2+1)\Rightarrow 3(3x-2)z^2+x=0$$$$\Rightarrow$$$$\frac{583929x{(x-\frac{1}{3})}^2}{{(405x^4+18x^2+1)}^2}(x^6+\frac{50}{267}{x}^5+\frac{31}{801}{x}^4+\frac{4}{801}{x}^3+\frac{7}{7209}{x}^2+\frac{2}{21627}x+\frac{1}{64881})=0$$$$x=0\wedge x=\frac{1}{3}\mathrm{is}\mathrm{obvious}\mathrm{because}\mathrm{of}\mathrm{the}\mathrm{first}$$$$\mathrm{attempt}\mathrm{solving}\mathrm{in}\mathbb{R}$$$$\mathrm{now}\mathrm{we}\mathrm{must}\mathrm{approximate}\mathrm{with}\mathrm{a}\mathrm{good}$$$$\mathrm{calculator}$$

Commented by i jagooll last updated on 11/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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