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Question Number 75845 by behi83417@gmail.com last updated on 18/Dec/19

$$\{\begin{array}{l}\mathbf{x}+\mathbf{yz}={\mathbf{x}}^2\\ \mathbf{y}+\mathbf{xz}={\mathbf{y}}^2\\ \mathbf{z}+\mathbf{xy}={\mathbf{z}}^2\end{array}\mathbf{solve}\mathbf{for}\mathbf{x},\mathbf{y},\mathbf{z}.$$

Commented by mr W last updated on 18/Dec/19

$$x=y=z=0$$

Commented by behi83417@gmail.com last updated on 18/Dec/19

$$\mathrm{thank}\mathrm{you}\mathrm{proph}:\mathrm{mrW}.$$$$\mathrm{another}\mathrm{solutions}?$$

Answered by MJS last updated on 18/Dec/19

$$\mathrm{trivial}\mathrm{solution}x=y=z=0$$$$\mathrm{with}x,y,z\ne 0$$$$\left(1\right)x+yz=x^2\Rightarrow z=\frac{x(x-1)}{y}$$$$\left(2\right)\frac{x^3-x^2+y^2}{y}=y^2$$$$\left(3\right)\frac{x(x+y^2-1)}{y}=\frac{x^2{(x-1)}^2}{y^2}$$$$\left(2\right)(x-y)(x^2+y^2+xy-x-y)=0$$$$\left(3\right)x(x-y-1)(x^2+y^2+xy-x-y)=0$$$$\Rightarrow \mathrm{one}\mathrm{out}\mathrm{of}x,y,z=-\frac{1}{3}\mathrm{and}\mathrm{the}\mathrm{other}$$$$\mathrm{two}=\frac{2}{3}\mathrm{or}$$$$\mathrm{two}\mathrm{out}\mathrm{of}x,y,z\mathrm{are}\mathrm{the}\mathrm{conjugated}$$$$\mathrm{pair}\frac{-t+1\pm \sqrt{(1-t)(3t+1)}}{2}\mathrm{with}t\mathrm{being}$$$$\mathrm{the}\mathrm{third}\mathrm{one};-\frac{1}{3}⩽t⩽1\mathrm{for}\mathrm{real}\mathrm{solutions}$$

Commented by behi83417@gmail.com last updated on 18/Dec/19

$$\mathrm{thanks}\mathrm{in}\mathrm{advance}\mathrm{dear}\mathrm{proph}.$$

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