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Question Number 180603 by mr W last updated on 14/Nov/22

$$findtherealsolutionoffollowing$$$$equationsystem:$$$${\mathit{x}}^2+\mathit{x}\mathit{y}+{\mathit{y}}^2=\mathit{p}$$$${\mathit{y}}^2+\mathit{y}\mathit{z}+{\mathit{z}}^2=\mathit{q}$$$${\mathit{z}}^2+\mathit{z}\mathit{x}+{\mathit{x}}^2=\mathit{r}$$$$withp,q,r>0$$

Commented by mr W last updated on 14/Nov/22

$$thegeneralsolutionis$$$$\lambda =\sqrt{(\sqrt{p}+\sqrt{q}+\sqrt{r})(-\sqrt{p}+\sqrt{q}+\sqrt{r})(\sqrt{p}-\sqrt{q}+\sqrt{r})(\sqrt{p}+\sqrt{q}-\sqrt{r})}$$$$x=\pm \frac{\sqrt{6}\lambda +3\sqrt{2}(p-q+r)}{6\sqrt{p+q+r+\sqrt{3}\lambda }}$$$$y=\pm \frac{\sqrt{6}\lambda +3\sqrt{2}(p+q-r)}{6\sqrt{p+q+r+\sqrt{3}\lambda }}$$$$z=\pm \frac{\sqrt{6}\lambda +3\sqrt{2}(-p+q+r)}{6\sqrt{p+q+r+\sqrt{3}\lambda }}$$

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