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Question Number 139842 by I want to learn more last updated on 01/May/21

Commented by MJS_new last updated on 01/May/21

$$\mathrm{no}\mathrm{real}\mathrm{solution}$$

Answered by mr W last updated on 01/May/21

$${(x+y+z)}^2=x^2+y^2+z^2+2(xy+yz+zx)$$$$\Rightarrow xy+yz+zx=\frac{7^2-9}{2}=20$$$$x+y+z=7$$$$xyz=5$$$$\Rightarrow x,y,zarerootsoffollowingeqn.:$$$$t^3-7t^2+20t-5=0$$$$lett=s+\frac{7}{3}$$$$\Rightarrow s^3+\frac{11}{3}s+\frac{439}{27}=0$$$$\sqrt{\mathrm{\Delta }}=\sqrt{{\left(\frac{11}{9}\right)}^3+{\left(\frac{439}{54}\right)}^2}=\frac{\sqrt{2445}}{6}$$$$s_1=\sqrt[3]{\frac{\sqrt{2445}}{6}-\frac{439}{54}}-\sqrt[3]{\frac{\sqrt{2445}}{6}+\frac{439}{54}}$$$$\Rightarrow t_1=\frac{7}{3}+\sqrt[3]{\frac{\sqrt{2445}}{6}-\frac{439}{54}}-\sqrt[3]{\frac{\sqrt{2445}}{6}+\frac{439}{54}}$$$$\approx 0.275524$$$$s_{2,3}=\frac{1}{2}(\sqrt[3]{\frac{\sqrt{2445}}{6}+\frac{439}{54}}-\sqrt[3]{\frac{\sqrt{2445}}{6}-\frac{439}{54}})\pm \frac{\sqrt{3}}{2}(\sqrt[3]{\frac{\sqrt{2445}}{6}+\frac{439}{54}}+\sqrt[3]{\frac{\sqrt{2445}}{6}-\frac{439}{54}})i$$$$\Rightarrow t_{2,3}=\frac{7}{3}+\frac{1}{2}(\sqrt[3]{\frac{\sqrt{2445}}{6}+\frac{439}{54}}-\sqrt[3]{\frac{\sqrt{2445}}{6}-\frac{439}{54}})\pm \frac{\sqrt{3}}{2}(\sqrt[3]{\frac{\sqrt{2445}}{6}+\frac{439}{54}}+\sqrt[3]{\frac{\sqrt{2445}}{6}-\frac{439}{54}})i$$$$\approx 3.362238\pm 2.615837i$$$$$$$$(x,y,z)\in (t_1,t_2,t_3)$$

Commented by I want to learn more last updated on 01/May/21

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by I want to learn more last updated on 01/May/21

$$\mathrm{Question}\mathrm{sir}.$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{formular}\mathrm{to}\mathrm{compute}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{t}\mathrm{sir}.\mathrm{Both}\mathrm{real}\mathrm{and}\mathrm{complex}.$$

Commented by mr W last updated on 02/May/21

$$Q93963$$

Commented by mr W last updated on 02/May/21

$$Q89687$$

Commented by I want to learn more last updated on 02/May/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by Tawa11 last updated on 23/Jul/21

$$\mathrm{great}$$

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