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I have the following equation system with $x,y,z \in \mathbb{R^{*}}$

\begin{equation*} { x+y+z = 1\\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z} =1\\ xy+yz+zx = -4 } \end{equation*}

My question is: How many solutions there are?

Iuli
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    Hint: the second equation gives $,xyz = -4,$, then you know $,x+y+z, xy+xz+yz, xyz,$ so $x,y,z$ are the roots of a certain cubic equation. – dxiv May 06 '22 at 17:56

1 Answers1

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As @dxiv pointed out in his comment above, multiplying the second equation by $xyz$ yields

$yz + xz + xy = xyz $

But from the third equation, we know that the left hand side is $-4$, hence $xyz = -4$

Now consider the cubic polynomial

$f(t) = (t - x)(t - y) (t - z) = t^3 - (x + y + z) t^2 + (xy + xz + yz) t - xyz $

With the given information,

$f(t) = t^3 - t - 4 t + 4 $

which factors into

$ f(t) = t (t^2 - 1) - 4 (t - 1) = (t - 1) ( t^2 + t - 4 ) $

whose roots are to a permutation, $ t = 1, \dfrac{1}{2} (-1 \pm \sqrt{17} ) $

Since $3! = 6 $ there are $6$ solutions.

Hosam Hajeer
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