Finding Solutions to a Symmetric Non-Linear Equation (for Some Cases Beside $x = y = z$) 
Find all values $x,y,z$ (whether real or complex) such that :

$2x-2y+z^{-1} = 2022^{-1}\\ 2z-2x+y^{-1}=2022^{-1}\\ 2y-2z + x^{-1} = 2022^{-1}$


I know that for case when $x=y=z$ i can easily find : $x=y=z = 2022$ (which seems to be the only real solution), What i'm struggling with is finding $x,y,z $ for cases beside $x = y = z$.
Attempt :

*

*$\displaystyle B_1+B_2+B_3 : x^{-1}+y^{-1}+z^{-1} = 3\cdot 2022^{-1}$


*$\displaystyle B_1-B_2+B_3 : 4(x-y) + 2z^{-1}-y^{-1}=2022^{-1} $


*$\displaystyle B_1+B_2 : 2(z-y) =2\cdot2022^{-1} $
It seems like no matter which row i'm operating i can never find equation in form of :
$ F_1(x,y,z)\cdot (x-y) = F_2(x,y,z) \cdot (x-y)  \\ F_3(x,y,z)\cdot (x-z) = F_4(x,y,z) \cdot (x-z) \\ F_5(x,y,z)\cdot (y-z) = F_6(x,y,z)\cdot (y-z)  $
My question : how do i turn the early equation into such equation above
so that i wouldn't need to find for $x/y/z$ manually (in a longer & standard way).
I need an assistance with this one, any help is appreciated
 A: Multiplying the first equation by $z$, the second by $y$ and the third by $x$ we obtain a system of polynomial equations
\begin{align*}
4044xz - 4044yz - z + 2022 & = 0,\\
- 4044xy + 4044yz - y + 2022 & = 0, \\
4044xy - 4044xz - x + 2022 & = 0.
\end{align*}
Adding the equations we obtain
$$
x+y+z=6066
$$
Substituting $z$ into the other equations and taking the resultant, we obtain
$$
(16353936x^3 - 99202975776x^2 - x + 674)(x - 2022)=0
$$
This yields either $x=y=z=2022$ or $x$ is one of the roots of the cubic. These are
\begin{align*} 
x & =6066.0000000000089603046877948096040149,\\
x & = 0.000082426635809989490801557561241189815030,\\
x & =-0.000082426644770294178596367165256134654133
\end{align*}
A: To simplify the calculations, doing the same as @Dietrich Burde did in his answer, let $a=\frac 1{2022}$ and you end with the cubic
$$12 a x^3-36 x^2-3 a^3 x+a^2=0$$
which has three real roots since
$$\Delta=1296 a^2 \left(a^4+12\right)^2 \quad > 0~~\forall a$$ Then the three roots are given by
$$x_k=\frac 1 a +\frac{\sqrt{3\left(a^4+12\right) }}{3 a}\cos \left(\frac{1}{3} \left(2 \pi  k-\cos
   ^{-1}\left(\sqrt{\frac{12}{a^4+12}}\right)\right)\right) \qquad (k=0,1,2)\tag 1$$ Using $a=\frac 1{2022}$, you obtain the results already given by @Dietrich Burde but the two small roots are not identical in absolute value; in this case, they differ by $8.96\times 10^{-12}$.
In fact, if $a$ is small, a Taylor expansion gives
$$x_1+x_2=-\frac{2 a^3}{27}+O\left(a^7\right)$$
Edit
Using $(1)$, for small values of $a$, the series expansions are
$$x_0=\frac{3}{a}+\frac{2 a^3}{27}-\frac{7 a^7}{4374}+O\left(a^{11}\right)$$
$$x_1=\frac{a}{6}-\frac{a^3}{27}+\frac{a^5}{486}+\frac{7 a^7}{8748}-\frac{5
   a^9}{52488}+O\left(a^{11}\right)$$
$$x_2=-\frac{a}{6}-\frac{a^3}{27}-\frac{a^5}{486}+\frac{7 a^7}{8748}+\frac{5a^9}{52488}+O\left(a^{11}\right)$$
