Is there a simpler approach to these system of equations? I recently came across the following system of equations:
$$x + y + z = 1 \\
x^2 + y^2 + z^2 = 2  \\
x^3 + y ^3 + z^3 = 3$$  
And I have two questions:
One, is there a way to prove or disprove whether there is a solution for this particular set of equations? Furthermore, is there a way to expand the proof for a more generalized set of equations, that is for this set :
$$x + y + z = 1 \\ x^2 + y^2 + z^2 = 2 \\ ...\\x^n + y^n + z^n = n$$
Two, is there a simpler approach for the prior solution set than substitution? As of now, I'm not getting anywhere with this method. I end up getting into a long-winded series of substitution and isolation that yields something like $z = z$ or $1 = 1$.
Sorry if it's a repeat question; I couldn't exactly find an accurate way to search for the equations.

We will make this into a cubic equation in terms of $z$.  We have $x+y = 1-z$ and $x^2 + y^2 = 2-z^2$.
Factor: $$x^3 + y^3 = ( x + y ) ( x^2 - xy + y^2 ) = ( 1-z ) ( 2-z^2 - xy )$$
$$2xy = ( x + y )^2 - ( x^2 + y^2 ) = ( 1-z )^2 - ( 2 - z^2 ) = 2z^2 - 2z - 1$$
So our third equation becomes (multiplying by $2$):
$$( 1 - z ) ( -4z^2 + 2z + 5 ) + 2z^3 = 6$$
Multiplying both sides by $2$ and expanding the polynomial, 
$$6z^3 - 6z^2 - 3z -1 = 0$$
This will have $3$ roots (in the complex numbers), which will be solutions for $x$, $y$, and $z$ (by symmetry).
