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#1 h Apr 22, 2025, 4:36 PM

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Solve the system of equations
$\begin{cases} x - y z = 1,\\[2pt] y - z x = 2,\\[2pt] z - x y = 4. \end{cases}$

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#2 h Apr 22, 2025, 6:14 PM

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#3 h Apr 22, 2025, 8:26 PM

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Plugging in $x=yz+1$ to the second and third equation gives $y-yz^2-z=2$ and $z-y^2z-y=4$ .

Rearrange the former of these equations to $y=\frac{z+2}{1-z^2}$ and plug it in the latter to get $z-\left(\frac{z+2}{1-z^2}\right)^2z-\frac{z+2}{1-z^2}=4$ .

Now simplify:

$z-\left(\frac{z+2}{1-z^2}\right)^2z-\frac{z+2}{1-z^2}=4\implies z(1-z^2)^2-(z+2)^2z+(z+2)(z^2-1)-4(1-z^2)^2=0\implies z^5-4z^4-2z^3+6z^2-4z-6=0$

Starting over and going through similar steps gives $y^5-2y^4-2y^3-16y-6=0$ and $x^5-x^4-2x^3-6x^2-19x-9=0$ .

Each of these quintics is irreducible with three real roots and as a consequence have Galois Group $S_5$ . So they can't be solved closed form by radicals.

The approximate solutions.

Real:
$(-1.50383,3.18296,-0.786634)$
$(-0.563382,-0.371416,4.20925)$
$(3.06006,-1.70257,-1.20997)$

Complex:
$(0.0036+1.8632i,0.4455+1.6681i,0.89368+0.83604i)$
$(0.0036-1.8632i,0.4455-1.6681i,0.89368-0.83604i)$

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