How can I solve this system of equations?? I have this system of equations and I need values for $c_1, c_2, c_3, x_1, x_2, x_3$.
\begin{align*}
c_1 + c_2 + c_3 &= 2\\
c_1x_1 + c_2x_2+ c_3x_3 &= 0\\
c_1x_1^2 + c_2x_2^2+c_3x_3^2 &= \frac{2}{3}\\
c_1x_1^3 + c_2x_2^3+c_3x_3^3 &= 0\\
c_1x_1^4 + c_2x_2^4 + c_3x_3^4 &= \frac{2}{5}\\
c_1x_1^5 + c_2x_2^5+c_3x_3^5 &= 0
\end{align*}
I have attempted using matrix but was unsuccessful. How can I obtain values for $x_i$ and $ c_i$?
 A: Rewriting the system as
\begin{align} 
\sum_{j=1}^3 c_j\,x_j^i&=v_i,\quad i=0,\dots,5
\tag{1}\label{1}
,
\end{align}  
we can apply Prony's method 
as follows.
Solve the linear system
\begin{align} 
\left[\begin{matrix}
v_0 & v_1 & v_2 \\
v_1 & v_2 & v_3 \\
v_2 & v_3 & v_4 
\end{matrix}\right]
\cdot
\left[\begin{matrix}
a_0 \\ a_1 \\ a_2
\end{matrix}\right]
&=
\left[\begin{matrix}
 v_3 \\ v_4 \\ v_5
\end{matrix}\right]
\end{align}  
for $a_0,a_1,a_2$.
The roots of polynomial
\begin{align} 
x^3-a_2\,x^2-a_1\,x-a_0
\end{align}
would be the triple $x_1,x_2,x_3$.
Given that, the solution of another linear system
\begin{align} 
\left[\begin{matrix}
1&1&1 \\
x_1&x_2&x_3 \\
x_1^2&x_2^2&x_3^2 
\end{matrix}\right]
\cdot
\left[\begin{matrix}
 c_1 \\ c_2 \\ c_3
\end{matrix}\right]
&=
\left[\begin{matrix}
 v_0 \\ v_1 \\ v_2
\end{matrix}\right]
\end{align}
for $c_1,c_2,c_3$ completes the answer.
In numbers we have
\begin{align}
\left[\begin{matrix}
 2 & 0 & \tfrac23 \\
 0 & \tfrac23 & 0 \\
 \tfrac23  & 0 & \tfrac25  
\end{matrix}\right]
\cdot
\left[\begin{matrix}
 a_0 \\ a_1 \\ a_2
\end{matrix}\right]
&=
\left[\begin{matrix}
 0 \\ \tfrac25 \\ 0
\end{matrix}\right]
,\\
\end{align}
\begin{align}
a_0&=0,\quad a_1=\tfrac35.\quad a_2=0
,\\
x^3-\tfrac35\,x&=0
,\\
x_1&=0,\quad x_2=\tfrac{\sqrt{15}}5
,\quad x_3=-\tfrac{\sqrt{15}}5
.
\end{align}
The system for $c_j$:
\begin{align} 
\left[\begin{matrix}
 1&1&\phantom{-}1 \\
 0& \tfrac{\sqrt{15}}5 & -\tfrac{\sqrt{15}}5 \\
 0& \tfrac35 & \phantom{-}\tfrac35 
\end{matrix}\right]
\cdot
\left[\begin{matrix}
 c_1 \\ c_2 \\ c_3
\end{matrix}\right]
&=
\left[\begin{matrix}
 2 \\ 0 \\ \tfrac23
\end{matrix}\right]
,\\
\end{align}
which results in
\begin{align} 
c_1&=\tfrac89,\quad 
c_2= c_3=\tfrac59
.
\end{align}
