Question Number 180603 by mr W last updated on 14/Nov/22
$${find}\:{the}\:{real}\:{solution}\:{of}\:{following} \\ $$$${equation}\:{system}: \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{xy}}+\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{p}} \\ $$$$\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{yz}}+\boldsymbol{{z}}^{\mathrm{2}} =\boldsymbol{{q}} \\ $$$$\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{zx}}+\boldsymbol{{x}}^{\mathrm{2}} =\boldsymbol{{r}} \\ $$$${with}\:{p},{q},{r}>\mathrm{0} \\ $$
Commented by mr W last updated on 14/Nov/22
$${the}\:{general}\:{solution}\:{is} \\ $$$$\lambda=\sqrt{\left(\sqrt{{p}}+\sqrt{{q}}+\sqrt{{r}}\right)\left(−\sqrt{{p}}+\sqrt{{q}}+\sqrt{{r}}\right)\left(\sqrt{{p}}−\sqrt{{q}}+\sqrt{{r}}\right)\left(\sqrt{{p}}+\sqrt{{q}}−\sqrt{{r}}\right)} \\ $$$${x}=\pm\frac{\sqrt{\mathrm{6}}\lambda+\mathrm{3}\sqrt{\mathrm{2}}\left({p}−{q}+{r}\right)}{\mathrm{6}\sqrt{{p}+{q}+{r}+\sqrt{\mathrm{3}}\lambda}} \\ $$$${y}=\pm\frac{\sqrt{\mathrm{6}}\lambda+\mathrm{3}\sqrt{\mathrm{2}}\left({p}+{q}−{r}\right)}{\mathrm{6}\sqrt{{p}+{q}+{r}+\sqrt{\mathrm{3}}\lambda}} \\ $$$${z}=\pm\frac{\sqrt{\mathrm{6}}\lambda+\mathrm{3}\sqrt{\mathrm{2}}\left(−{p}+{q}+{r}\right)}{\mathrm{6}\sqrt{{p}+{q}+{r}+\sqrt{\mathrm{3}}\lambda}} \\ $$