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.

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.
