<s>[INST] 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.
 [/INST] 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.
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