Analytical Solution to a nonlinear system of equations Is it possible to analytically solve for x,y,z in the following system, and if so how would one go about it?
$a_{1} (y - 2x + z) - x + 2x^{3} - a_{2}x = 0$
$a_{1} (z - 2y + x) - y + 2y^{3} - a_{2}y = 0$
$a_{1} (x - 2z + y) - z + 2z^{3} - a_{2}z = 0$
I've tried numerous methods such as cyclic permutations of the three equations and substitution but none seem to work, hence any advice would be greatly appreciated.
 A: Let $\,s = x+y+z\,$ then the first equation can be written as:
$$
\begin{align}
0 &= a_{1} (y - 2x + z) - x + 2x^{3} - a_{2}x
\\ &= a_{1} (s - 3x) - x + 2x^{3} - a_{2}x
\\ &= 2x^{3} - (3 a_1 + a_2 + 1) x + a_{1} s
\end{align}
$$
The same equation is satisfied by $\,y\,$ and $\,z\,$. Assuming $\,x,y,z\,$ are distinct, they are the roots of the same depressed cubic, so their sum must be $\,s = 0\,$, and the equations reduce to:
$$
2t^{3} - (3 a_1 + a_2 + 1) t = 0 \quad\quad \Big|\quad t = x, y, z
$$
Therefore the non-trivial distinct solutions of the system are permutations of:
$$
x = \sqrt{\frac{3a_1+a_2+1}{2}} \;,\quad y = - \sqrt{\frac{3a_1+a_2+1}{2}}\;, \quad z = 0
$$
This leaves the case where pairs of $\,x,y,z\,$ are equal to be solved separately.

[ EDIT ] The following covers the remaining case where two solutions are equal. Assume WLOG that $\,y=z\,$, then the last equation becomes redundant, and the system reduces to:
$$
\begin{cases}
\begin{align}
2x^{3} - (a_2+1) x - 2a_{1} (x-y) &= 0
\\ 2y^{3} - (a_2 + 1)y + a_{1} (x-y) &= 0
\end{align}
\end{cases}
$$
It can be seen "by inspection" that a solution is $\,x=y\,$ where $\,x\,$ is a root of $\,2x^{3} - (a_2+1) x = 0\,$, which gives the identical triplets $\,x = y = z \in \left\{0, \pm \sqrt{\dfrac{a_2+1}{2}}\right\}\,$ as solutions.
To find all solutions, the first equation gives $\,y\,$ as a cubic in $\,x\,$, then substituting into the second equation gives an equation of degree $\,9\,$ in $\,x\,$ alone. We know from the previous step that $\,2x^{3} - (a_2+1) x\,$ must be a factor, so that leaves a sextic in $\,x\,$, which in the end turns out to be a cubic in $\,x^2\,$ (courtesy WA):
$$
x^2 (a_2 - 2 x^2 + 1)^2 + 6 a_1 x^2 (a_2 - 2 x^2 + 1) - 2 a_1^2 (a_2 - 6 x^2 + 1) - 6 a_1^3 = 0
$$
A: Notice the symmetry in the equations.  Adding them all together, we find that
$$-(x+y+z)+2(x^3+y^3+z^3)-a_2(x+y+z) = 0$$
$$2(x^3+y^3+z^3) = (1+a_2)(x+y+z).$$
Now, we utilize the convenient factored form:
$$x^3+y^3+z^3 - (x+y+z) = (x+y+z)(x+y+z+1)(x+y+z-1),$$
to write the above equation as
$$(x+y+z)(x+y+z+1)(x+y+z-1) = \frac{a_2-1}{2}(x+y+z).$$
Assuming $x+y+z \ne 0$,
$$(x+y+z+1)(x+y+z-1) = \frac{a_2-1}{2}.$$
Thus,
$$z = \pm \sqrt{\frac{a_2+1}{2}}-x-y.$$
Notice that if we substitute this into the original first equation, we find
$$a_1(\sqrt{\frac{a_2+1}{2}}-3x)+2x^3-a_2x = 0.$$
This is a cubic in $x$, which you can solve.
Proceed in a similar fashion in the second original equation to find a cubic in terms only of $y$.  Finally, solve $z = \pm \sqrt{\frac{a_2+1}{2}}-x-y$ to find $z$.
Note, you will also have to check for solution when $x+y+z = 0$, but I think this will at least give you a good start.
By using $s = x+y+z$ and looking at first part of dxiv's answer, this case is very simple.
