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It seems that $x=y=z=1$ is the only solution, but I can't find how to prove it. I've tried defining $a=x-1; b=y-1; c=z-1$, but with no success. Any ideas?

Geometrically, each equation defines a 3D surface. All 3 surfaces are identical, except they are oriented along different axes. We're looking for their intersection.

MiniQuark
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  • I checked using wolfram alpha and all other solutions are have imaginary values. – M.L Jun 24 '21 at 05:29

2 Answers2

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The system can be written as:

$$ \begin{align} \begin{cases} \;\; (x-1) &+\;\; (y-1)\,(y+1) &+\;\; (z-1)\,(z^2+z+1) &= 0 \\ \;\; (x-1)\,(x^2+x+1) &+\;\;(y-1) &+\;\; (z-1)\,(z+1) &= 0 \\ \;\; (x-1)\,(x+1) &+\;\;(y-1)\,(y^2+y+1) &+\;\; (z-1) &= 0 \end{cases} \end{align} $$

Regarding it as a linear homogeneous system in $x-1, y-1, z-1$, its determinant is:

$$ \Delta = \begin{vmatrix} 1 &y+1 &z^2+z+1 \\ x^2+x+1 &1 &z+1 \\ x+1 &y^2+y+1 &1 \end{vmatrix} $$

After routine calculations courtesy WA, and using that $x,y,z \gt 0\,$:

$$ \begin{align} \Delta &= x^2 y^2 z^2 + x^2 y^2 z + x^2 y^2 + x^2 y z^2 + x^2 y z + x^2 z^2 + x^2 z + x y^2 z^2 \\ &\;\;\;\; + x y^2 z + x y^2 + x y z^2 + 2 x y z + x y + x z + y^2 z^2 + y z^2 + y z \\ &\gt 0 \end{align} $$

It follows that the only solution is $x-1=y-1=z-1=0\,$.

dxiv
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  • No, you can’t solve the system of equations as if it were a linear system like that. The parameters depend on the variables $x,y,z$ – NN2 Jun 24 '21 at 06:53
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    @NN2 Yes, you can. Define the auxiliary variables $a=x-1, b=y-1, c=z-1$ (like the OP did in the question), then you have a linear homogeneous system in $a,b,c$ with coefficients that are polynomials in $x,y,z$. If the determinant of the system is proved to be non-$0$ (as is the case here) then the only solution is $a=b=c=0$. – dxiv Jun 24 '21 at 06:55
  • @dxiv sorry that I'm looking at your answer only now but what about : $\begin{cases}x^2+y^2=0\x^2+y^2=0\end{cases}$ there is obviously a single solution whereas determinant is $\left|\begin{array} 0x&y\x&y\end{array}\right|=0$ – Olivier Mar 14 '22 at 07:03
  • @Olivier That is true, but unrelated. Point is that if the determinant is $\ne 0$ then the trivial solution is the only solution. What happens if the determinant $=0$ is of no interest and no consequence here. – dxiv Mar 14 '22 at 07:41
  • Ok @dxiv. Thank you. – Olivier Mar 14 '22 at 13:05
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Wolfram Mathematica confirms that there is only one solution:

Solve[{x + y^2 + z^3 == 3, y + z^2 + x^3 == 3, z + x^2 + y^3 == 3}, {x, y, z}, PositiveReals]

{{x -> 1, y -> 1, z -> 1}}