Solving $x+y+z=4$, $x^2+y^2+z^2=14$, $x^3+y^3+z^3=34$ 
Solve the system $$\begin{equation} \label{equation1}
\begin{split}
x+y+z=4 \\
     x^2+y^2+z^2=14 \\
x^3+y^3+z^3=34
\end{split}
\end{equation}$$

My work:
I found out that $$xy+yz+xz=1$$ and $$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=22$$
After this I'm stuck. Any help is greatly appreciated.
EDIT This not a duplicate. I'm looking for a detailed solution and not a solution just by inspection. Also I thought of a new idea. Maybe e should consider a cubic polynomial whose roots are $x,y,z$

There is always the way of a direct computation. Substituting $z=4-x-y$, the other two equations are $f(x,y)=0$ and $g(x,y)=0$, and the resultant of $f$ and $g$ with respect to $y$ yields
$$
(x + 1)(x - 2)(x - 3)=0.
$$
So we have $x=-1$ or $x=2$ or $x=3$. To be more precise,
\begin{align}
f(x,y)& =x^2 + xy - 4x + y^2 - 4y + 1,\\
g(x,y)& = - x^2y + 4x^2 - xy^2 + 8xy - 16x + 4y^2 - 16y + 10.
\end{align}
