How to solve this simultaneous equation of $3$ variables. I've stuck in this equation system. No clue how to start ?
$$\begin{eqnarray}
x+y+z &=&a+b+c\tag{1} \\
ax+by+cz &=&a^{2}+b^{2}+c^{2}\tag{2} \\
ax^{2}+by^{2}+cz^{2} &=&a^{3}+b^{3}+c^{3}\tag{3}
\end{eqnarray}$$
Find the value of $x,y,z$ is in the form of $a,b$ and $c$.
I want to know steps of solution.


I'm supposed to solve it using simultaneous eqn method like
  elimination,substitution,cross-multiplication .NO knowledge of
  matrices

By inspection we see that $(x,y,z)=(a,b,c)$ is a solution of the given system
$$ 
\begin{array}{l}
\text{Eq. 1}\qquad x+y+z=a+b+c \\ 
\text{Eq. 2}\qquad ax+by+cz=a^{2}+b^{2}+c^{2} \\ 
\text{Eq. 3}\qquad ax^{2}+by^{2}+cz^{2}=a^{3}+b^{3}+c^{3}.
\end{array}
 \tag{0}$$
The other solution can be found as follows. Solve Eq. 1 for $z$. Multiply original Eq. 1 by $a$, subtract Eq. 2 and solve for $z$. This results in
$$
\begin{array}{l}
z=a+b+c-x-y,
\end{array}
 \tag{1}$$
$$
\begin{array}{l}
z=\frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c} .
\end{array}
 \tag{2}$$
Equate the right hand sides of $(1)$ and $(2)$
$$
\begin{array}{l}
\frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c}=a+b+c-x-y, 
\end{array}
 \tag{3}$$
and solve for $x$ 
$$
x=\frac{c-b}{a-c}y+\frac{-ac+b^{2}+a^{2}-bc}{a-c}. 
\tag{4}$$
Substitute $x,z$ in $(0)$, Eq. 3, using $(4)$ and $(2)$
$$
a\left( \frac{c-b}{a-c}y+\frac{-ac+b^{2}+a^{2}-bc}{a-c}\right)
^{2}+by^{2}+c\left( \frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c}\right)
^{2}=a^{3}+b^{3}+c^{3}.
\tag{5}$$
Solving for $y$ we get$^1$ the solution $y=b$ and the solution
$$y=\frac{B}{D},\tag{6}$$
where
$$
B=-2a^{3}c+2a^{3}b-a^{2}b^{2}-a^{2}bc+4a^{2}c^{2}-2acb^{2}-2ac^{3}+ab^{3}-abc^{2}+2bc^{3}+b^{3}c-c^{2}b^{2}$$
$$
D=a^{2}c+ac^{2}-6abc+a^{2}b+bc^{2}+ab^{2}+b^{2}c
$$
Finally substitute $y=b$ and $y=B/D$ in $(4)$ and $(2)$. We get the solutions $(x,z)=(a,c)$, and
$$(x,z)=\left(\frac{A}{D},\frac{C}{D}\right),\tag{7}$$
where
$$
A=a^{3}c+a^{3}b-2a^{2}bc-a^{2}b^{2}-a^{2}c^{2}+2ac^{3}-abc^{2}+2ab^{3}-acb^{2}-2bc^{3}-2b^{3}c+4c^{2}b^{2}
$$
$$
C=2a^{3}c-2a^{3}b+4a^{2}b^{2}-a^{2}c^{2}-a^{2}bc-acb^{2}-2ab^{3}+ac^{3}-2abc^{2}+bc^{3}+2b^{3}c-c^{2}b^{2}.
$$
So the two solutions of $(0)$ are: $$(x,y,z)=(a,b,c)\qquad\text{and }\qquad(x,y,z)=\left(\frac{A}{D},\frac{B}{D},\frac{C}{D}\right).$$
--
$^1$ Eq. $(5)$ is equivalent to
$$\begin{equation*}
\left( cb^{2}+c^{2}b+ac^{2}+ca^{2}+ab^{2}-6acb+ba^{2}\right) (y-b)\left( y-\frac{B}{D}\right) =0.
\end{equation*}$$
