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.

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.
