TITLE: Solve system of nonlinear equations using non-numerical method
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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
$$
0=z^2-4z+2=(z-(2+\sqrt2))(z-(2-\sqrt(2)).
$$