Does this system have infinitely many solutions. Because evident solution is $1,-1,1$ . And I reached till condition $xyz=-1$ with algebraic transformations. I got that $x^2+y^2+z^2+2(xy+yz+zx)=1$ and from the second equation got $xy+yz+zx=-1$. Then got from $(x^3+y^3+z^3)-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$, that $xyz=-1$. I tried to return with the shift $x=-1/yz$ back to equations, however, I cannot get concrete solutions, does anyone have an idea how to proceed?
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For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Aug 23 '23 at 14:19
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Thank you i will use that next time – Maca Aug 23 '23 at 14:21
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Let\begin{align}P(\lambda)&=(\lambda-x)(\lambda-y)(\lambda-z)\\&=\lambda^3-(x+y+z)\lambda^2+(xy+xz+yz)\lambda-xyz\\&=\lambda^3-\lambda^2-\lambda+1.\end{align}Then $x$, $y$, and $z$ are the roots of $P(\lambda)$. But\begin{align}P(\lambda)=0&\iff\lambda^3-\lambda^2-\lambda+1=0\\&\iff\lambda^2(\lambda-1)-(\lambda-1)=0\\&\iff(\lambda^2-1)(\lambda-1)=0\\&\iff(\lambda-1)^2(\lambda+1)=0.\end{align}Therefore, the solutions are$$(x,y,z)=(1,1,-1),\ (x,y,z)=(1,-1,1)\text{, and }(x,y,z)=(-1,1,1).$$
José Carlos Santos
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Sorry i dont get it how we related this equation with the system?- José Carlos Santos – Maca Aug 23 '23 at 14:41
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@lone student Are Vietas formulas not for the equations with one unknown value. – Maca Aug 23 '23 at 14:59
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@José Carlos Santos I get it how we got till the solution -1,1,1 but i dont get it how to explain solution of a system through p(/lambda) sorry i know i am sure dumb for this but can you explain me please – Maca Aug 23 '23 at 15:01
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@Maca The numbers that you are after are $x$, $y$, and $z$. But $P(\lambda)=(\lambda-x)(\lambda-y)(\lambda-z)$. Therefore, the numbers that you are after are the roots of $P(\lambda)$, which turn out to be $1$, $1$, and $-1$. – José Carlos Santos Aug 23 '23 at 15:11