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.
 A: 
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
$$\begin{equation*}
\left( cb^{2}+c^{2}b+ac^{2}+ca^{2}+ab^{2}-6acb+ba^{2}\right) (y-b)\left( y-\frac{B}{D}\right) =0.
\end{equation*}$$
A: Assume that $(a,b,c)\ne \lambda(1,1,1)$ and put $x:=a+u$, $\>y:=b+v$, $\>z:=c+w$. Then $(1)$ and $(2)$ amount to the system
$$\eqalign{\phantom{a}u+\phantom{b}v+\phantom{c}w&=0\cr a u+ bv+cw&=0\cr}$$
with the solution
$$u=t(b-c),\quad v=t(c-a),\quad w=t(b-c)\qquad(t\in{\mathbb R})\ .$$
Plug this into the equation $(3)$:
$$a(a+u)^2+b(b+v)^2+c(c+w)^2=a^3+b^3+c^3$$
and obtain a quadratic equation for $t$ with one obvious solution $t=0$ and a second solution
$$t={2(a-b)(b-c)(c-a)\over ab(a+b)+bc(b+c)+ca(c+a)-6abc}\ .\tag{4}$$
Cases where the denominator in $(4)$ is zero have to be dealt with separately.
A: This is a slightly interesting question. Let me try and provide some motivation for the ugly values that Americo calculated.
Let me assume that $a\neq b, b \neq c, c \neq a $. We'd deal with this issue separately.
Note that the first 2 equations give the equation on a plane. Hence, their intersection is a line, which we can calculate. The first 2 equations are equivalent to 
$$ \begin{array} {llll}
&&(x-a) &+ &(y-b) &+ &(z-c) &= 0\\
&a&(x-a) &+b&(y-b) & +c&(z-c) & = 0 \\ \end{array}$$
Since
$$ \begin{pmatrix} 1\\1\\1\\ \end{pmatrix} \times  \begin{pmatrix} a\\b\\c\\ \end{pmatrix} = \begin{pmatrix} c-b\\a-c\\b-a\\ \end{pmatrix}, $$
we know that the line is given by $\frac{x-a}{c-b} = \frac{ y-b}{a-c} = \frac{z-c}{b-a} = k $. (This is valid because the denominator is never 0.) This is equivalent to $x=a + k(c-b), y = b+k(a-c) , z = c+k(b-a)$.
Next, we are interested in when this line intersects the ellipsoid $ax^2 + by^2 + cz^2 = a^3 + b^3 + z^3 $, which happens in at most 2 places. (Of course, one place is $k=0$, which leads to the solution $x=a, y=b, z=c$.) Using the equation of the line above, we get that
$$\begin{array}{l l l l}
ax^2 &= a^3 & + 2a^2k(c-b) &+k^2 a(c-b)^2 \\
by^2 &= b^3 & + 2b^2k(a-c) & + k^2 b(a-c)^2   \\
cz^2 &= c^3 & + 2c^2k(b-a) & + k^2 c(b-a)^2  \\
\end{array} $$
Summing up these three equations, and using the last equation given, we get that
$$-2k(a-b)(b-c)(c-a) + k^2 (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b - 6abc) = 0. $$
One solution is $k=0$ as mentioned, and the other is 
$$ k = \frac{ 2 (a-b)(b-c)(c-a) } { a^2b+a^2c+b^2a+b^2c+c^2a+c^2b - 6abc} .$$
With this value of $k$, we get the answer that Americo calculated above.

Now, what happens when $a=b$ or $b=c$ or $c=a$? Interestingly, this forces there to be only 1 solution, namely $k=0$. You can also get this result by substituting $a=b$ into the value of $k$ and showing that the numerator is 0. The geometric interpretation is that the line becomes tangential to the ellipsoid.
Similarly, if $a=b=c$, then there is only 1 solution.
