Solve system of nonlinear equations using non-numerical method Is there any non-numerical method to solve this kind of system of nonlinear equations for $c_1, c_2, x_1, x_2$: 
$$c_1+c_2 = 1$$
$$c_1x_1+c_2x_2 = 1$$
$$c_1x_1^2+c_2x_2^2 = 2$$
$$c_1x_1^3+c_2x_2^3 = 6$$
I tried to solve this system, but I cannot find the efficient way. 
 A: Groebner Basis solution (using sympy):
from sympy import *

c1, c2, x1, x2 = symbols('c1 c2 x1 x2')

G = groebner([c1+c2-1,c1*x1+c2*x2-1,c1*x1**2+c2*x2**2-2,c1*x1**3+c2*x2**3-6])
print G
C2 = solve(G[3])

#solution #1
C1 = solve(G[2].subs(c2, C2[0]))[0]
X2 = solve(G[1].subs(c2, C2[0]))[0]
X1 = solve(G[0].subs(c2, C2[0]))[0]
print 'X1 = ', X1, ', ', 'X2 = ', X2, ', ', 'C1 = ', C1, ', ', 'C2 = ', C2[0]

#solution #2
C1 = solve(G[2].subs(c2, C2[1]))[0]
X2 = solve(G[1].subs(c2, C2[1]))[0]
X1 = solve(G[0].subs(c2, C2[1]))[0]
print 'X1 = ', X1, ', ', 'X2 = ', X2, ', ', 'C1 = ', C1, ', ', 'C2 = ', C2[1]

here is the output:
GroebnerBasis([-4*c2 + x1, 4*c2 + x2 - 4, c1 + c2 - 1, 8*c2**2 - 8*c2 + 1], x1, x2, c1, c2, domain='ZZ', order='lex')
X1 =  -sqrt(2) + 2 ,  X2 =  sqrt(2) + 2 ,  C1 =  sqrt(2)/4 + 1/2 ,  C2 =  -sqrt(2)/4 + 1/2
X1 =  sqrt(2) + 2 ,  X2 =  -sqrt(2) + 2 ,  C1 =  -sqrt(2)/4 + 1/2 ,  C2 =  sqrt(2)/4 + 1/2

A: Write $a = c_1$, $u = x_1$,
and $v = x_2$.
Since $c_2 = 1-c_1 = 1-a$,
$a u^k + (1-a)v^k = r_k$
for $k = 1, 2, 3$
where the $r_k = 1, 2, 6$ 
 as above.
From this,
$a(u^k-v^k)+v^k = r_k$,
or
$a = \frac{r_k-v^k}{u^k-v^k}$.
Equating for $k=1,2$,
$\frac{r_1-v}{u-v}
=\frac{r_2-v^2}{u^2-v^2}
$,
or
$u+v = \frac{r_2-v^2}{r_1-v} $,
or
$u = \frac{r_2-v^2}{r_1-v}-v $.
Squaring,
we also have
$ \frac{(r_2-v^2)^2}{(r_1-v)^2}
=(u+v)^2 
=u^2+2uv+v^2 $.
Equating for $k=1,3$,
$\frac{r_1-v}{u-v}
=\frac{r_3-v^3}{u^3-v^3}
$,
or
$u^2+uv+v^2 = \frac{r_3-v^3}{r_1-v}$.
Subtracting these two,
$uv =\frac{(r_2-v^2)^2}{(r_1-v)^2}- \frac{r_3-v^3}{r_1-v}$
so that
$v(\frac{r_2-v^2}{r_1-v}-v)
 =\frac{(r_2-v^2)^2}{(r_1-v)^2}- \frac{r_3-v^3}{r_1-v}$.
This is a quartic in $v$,
which can be solved by 
the usual methods.
I'll leave it at this for now.
Finding any errors are left as an exercise for the reader,
but the method should be correct.
A: With polynomial roots
$p(x)=(x-x_1)(x-x_2)(x-a)$ is a cubic polynomial with roots $x_1,x_1,a$ and  coefficients
$$p(x)=x^3-(x_1+x_2+a)x^2+(x_1a+x_2a+x_1x_2)x-x_1x_2a.$$
Forming the linear combination of the given equations with these coefficients to give $c_1p(x_1)+c_2p(x_2)$ on the left side results in
$$
0=6-2(x_1+x_2+a)+(x_1a+x_2a+x_1x_2)-x_1x_2a
$$
which has to hold independent of the value of $a$. Thus by isolating the coefficients of powers of $a$ the two equations
\begin{align}
0&=6-2(x_1+x_2)+x_1x_2\\
0&=-2+(x_1+x_2)-x_1x_2
\end{align}
result, and in consequence (also directly by setting $a=1$ and $a=2$)
\begin{align}
x_1+x_2&=4\\
x_1x_2&=2
\end{align}
so that $x_{1/2}$ are the solutions of
$$
0=z^2-4z+2=(z-2)^2-2
$$
i.e., $x_{1/2}=2\pm\sqrt2$.

With a difference equation
The given equations form on the left the first terms of the general solution
$$
u_n=c_1⋅x_1^n+c_2⋅x_2^n
$$
of a general second order linear homogeneous recurrence equation
$$
0=u_{n+2}+a_1⋅u_{n+1}+a_0⋅u_n
$$
with $x_{1/2}$ the roots of its characteristic equation.
The values $(u_0,u_2,u_2,u_3)=(1,1,2,6)$ on the right form the start of a solution of this linear recurrence allowing to determine its coefficients. Insertion of the two contained triples gives the equations
\begin{align}
0&=2+a_1⋅1+a_0⋅1\\
0&=6+a_1⋅2+a_0⋅1
\end{align}
with solution $a_1=-4$, $a_0=2$, and thus the characteristic equation giving the same quadratic polynomial
$$
0=z^2-4z+2=(z-(2+\sqrt2))(z-(2-\sqrt(2)).
$$
