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

Using exactly the same approach as this answer here:
$$x+y+z=3$$
$$xy+yz+zx=3$$
$$xyz=1$$
And we then conclude that these are the roots of a cubic polynomial. This restricts the maximum number of solutions to $3$.
How do we know or show that the number of solutions is maximum $3$?

The fundamental theorem of algebra states that any polynomial will have the same number of roots as the degree of the polynomial (counting repeated roots).
Looking at the last set of equations, we can use Viète's formulas to find the cubic which solves the original set of equations.
$x + y + z = \frac{-b}{a}$, $xy + xz + yz = \frac{c}{a}$, $xyz = \frac{-d}{a}$, therefore plugging in the values from the last set of equations gives you: $3a = -b, 3a = c, a = -d$.
Knowing that the general form of a cubic is $ax^3 + bx^2 + cx + d$ and using the relations, the cubic used to solve the last set of equations is $ax^3-3ax^2+3a-a=0$. Since $a \neq 0$, $x^3 - 3x^2 + 3x - 1 = 0$.
Therefore, the cubic is guaranteed to have $3$ roots (including repeated roots) because of the fundamental theorem of algebra and hence, the last set of equations.
