<s>[INST] 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.
 [/INST] $$(x+y+z)^2=9=x^2+y^2+z^2+2(xy+yz+zx)\implies xy+yz+zx=3$$
$$(x+y+z)^3=27=6xyz+3(x+y+z)(x^2+y^2+z^2)-2(x^3+y^3+z^3)=6xyz+27-6\implies xyz=1$$
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.
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