USAMO 1973 (Simultaneous Equations) 
Determine all the roots, real or complex, of the system of simultaneous equations (USAMO 1973/4)
$$x+y+z=3$$
$$x^2+y^2+z^2=3$$
$$x^3+y^3+z^3=3$$

Multiply equation I by 2 and subtract it from Equation II:
$$x^2-2x+y^2-2y+z^2-2z=-3$$
Complete the square for all three variables:
$$(x-1)^2+(y-1)^2+(z-1)^2=0$$
Since the RHS is zero, and the LHS has only perfect squares, by the trivial inequality, we must have:
$$(x-1)^2=0\Rightarrow x=1$$
$$(y-1)^2\Rightarrow y=1$$
$$(z-1)^2=0\Rightarrow z=1$$
Hence, the only solution for Equations I and II is
$$x=y=z=1$$
This satisfies Equation III as well. And hence, this is also the only solution to the overall system of equations.
1.
Would this be enough to get full marks?
Anything missing, or needed to be added?
2.
The complex roots is neither needed nor used anywhere. This was just used to (artificially?)  increase the "complexity" of the problem. Is this correct.

The nicer solutions have been posted already, so here is a different one, just for fun.
Moving the constant terms $\,3\,$ to the LHS, distributing and factoring, the system can be written as:
$$
\begin{cases}
  (x-1) & +\; (y-1) & +\; (z-1) & = 0 \\
  (x-1) \cdot (x+1) & +\; (y-1) \cdot (y+1) & +\; (z-1) \cdot (z+1) & = 0 \\
  (x-1) \cdot (x^2+x+1) & +\; (y-1) \cdot (y^2+y+1) & +\; (z-1) \cdot (z^2+z+1) & = 0
\end{cases}
$$
Considering this as a linear system in $\,x-1,y-1,z-1\,$ its determinant is:
$$
\left| 
\begin{matrix}
\;1 \;&\; 1 \;&\; 1 \; \\
\;x+1 \;&\; y+1 \;&\; z+1 \; \\
\;x^2+x+1 \;&\; y^2+y+1 \;&\; z^2+z+1 \; \\
\end{matrix}
\right|
$$
After the obvious manipulations, this reduces to a Vandermonde determinant in variables $\,x,y,z\,$, which is non-zero iff all variables are different. But that case leads to the contradiction that the trivial solution $\,x-1=y-1=z-1=0\,$ has all variables equal. So the determinant must be $\,0\,$ i.e. two of the variables be equal. Assuming $\,x=y\,$ for example, it is straightforward to verify that the only solution is, again, $\,x=y=z=1\,$.
