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Question Number 164077 by mathlove last updated on 13/Jan/22

Answered by cortano1 last updated on 13/Jan/22

$$\left(1\right)x+y+xy+1=4\Rightarrow (x+1)(y+1)=4$$$$\left(2\right)y+z+yz+1=6\Rightarrow (y+1)(z+1)=6$$$$\left(3\right)z+x+zx+1=8\Rightarrow (z+1)(x+1)=8$$$$\left(1\right)\times \left(2\right)\times \left(3\right)\Rightarrow (x+1)(y+1)(z+1)=\pm \sqrt{4\times 4\times 4\times 3}=\pm 8\sqrt{3}$$$$z+1=\pm 2\sqrt{3}$$$$x+1=\pm \frac{4\sqrt{3}}{3}$$$$y+1=\pm \sqrt{3}$$

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