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Question Number 176472 by mr W last updated on 19/Sep/22

$$solveforx,y,zwith$$$$x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=5$$

Answered by behi834171 last updated on 19/Sep/22

$$x+y+z=5$$$$x^2+x(y+z)=5xandsoon$$$$\Rightarrow \mathrm{\Sigma }{x}^2+2\mathrm{\Sigma }xy=5\mathrm{\Sigma }x$$$$\Rightarrow 5+2\mathrm{\Sigma }xy=5\times 5\Rightarrow \mathrm{\Sigma }\mathit{x}\mathit{y}=10$$$$\mathrm{\Sigma }{x}^2=5\Rightarrow x^3+x(y^2+z^2)=5xandsoon..$$$$\Rightarrow \mathrm{\Sigma }{x}^3+\mathrm{\Sigma }x(y^2+z^2)=5\mathrm{\Sigma }x\Rightarrow$$$$\Rightarrow \mathrm{\Sigma }{x}^3+\mathrm{\Sigma }xy(x+y)=5\mathrm{\Sigma }x\Rightarrow$$$$\Rightarrow \mathrm{\Sigma }{x}^3+\mathrm{\Sigma }xy(5-z)=5\mathrm{\Sigma }x\Rightarrow$$$$\Rightarrow \mathrm{\Sigma }{x}^3+5\mathrm{\Sigma }xy-3xyz=5\mathrm{\Sigma }x$$$$\Rightarrow 5+5\times 10-3xyz=5\times 5\Rightarrow \mathit{x}\mathit{y}\mathit{z}=10$$$$\mathit{s}\mathit{o}:x,y,z;aretherootsof:$$$${\mathit{t}}^3-5{\mathit{t}}^2+10\mathit{t}-10=0$$$$thishaveonlyonerealroot.$$

Commented by Tawa11 last updated on 20/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by mr W last updated on 19/Sep/22

$$thankssir!$$

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