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.
 A: $$(x+y+z)^2=9=x^2+y^2+z^2+2(xy+yz+zx)\implies xy+yz+zx=3$$
$$(x+y+z)^3=27=6xyz+3(x+y+z)(x^2+y^2+z^2)-2(x^3+y^3+z^3)=6xyz+27-6\implies xyz=1$$
Hence by Viète's formulas $x,y,z$ are the roots of $t^3-3t^2+3t-1=(t-1)^3$, so $x=y=z=1$ is the only solution.
Note that Viète's formulas are valid even in the complex plane; the specification of "complex" in the question precludes using real-line tricks like the $x^2\ge0$ inequality in your attempt.
A: As pointed out in the comments,

*

*This solution is completely wrong and worth 0 points. You incorrectly assumed that $x-1, y-1, z-1$ are real numbers.


*This is incorrect. Solving for "real roots" is very different from solving for "complex roots".

Claim: With $n$(=3 in this case) variables, Newton's identities tell us that the first $n$ power sums uniquely determine the $n$ elementary symmetric polynomials, which uniquely determines the  complex $x_i$ (up to permutation) via Vieta's formula.
One idea/line solution: Hence, this system has a unique solution (up to permutation of $x, y, z$), which we observe to be $(1, 1, 1)$.
Thus there is only 1 complex solution.

I leave it to you to verify the claim and unpack this statement.
It uses more machinery than is necessary to solve this problem, but showcases the underlying mathematics, which is why this is my favorite approach. I'm not saying that one approach is better than the other.
Parcly's solution does the work of finding these elementary symmetric polynomials, while I avoid calculating them by guessing the unique solution.
Parcly's approach is guaranteed to obtain the solution, while mine requires some luck.
A: 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, $\,x=y=z=1\,$.
