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 $x,y,z$

$x+y+z=4$
$x^2+y^2+z^2=14$
$xy+yz+zx=1$(you already established)....(1)
$x^3+y^3+z^3-3xyz+3xyz=34$ 
$\Rightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz=34$ 
$\Rightarrow 4(14-1)+3xyz=34$
So, $xyz=-6$....(2)
From (1) and (2)
$(1/x)+(1/y)+(1/z)=-(1/6)$
$\Rightarrow (x+y)/xy+(1/z)=-(1/6)$
$\Rightarrow (4-z)/(-6/z)+(1/z)=-(1/6)$
Simplifying we get
$z^3-4z^2+z+6=0$
$\Rightarrow (z-3)(z-2)(z+1)=0$
$\therefore, z=3,2,-1$
Put these values in $x+y+z=4$ and $xyz=-6$, you can easily get the values of $x$ and $y$.
There will be 6 solutions, permutations of
$(3,2,-1)$.
