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 $x,y,z$

We observe that,
$$xy=\frac{(4-z)^2-(14-z^2)}{2}=f(z)$$
and
$$\begin{align}(4-z)(14-z^2)=xy(x+y)+x^3+y^3\\
=f(z)\times (4-z)+34-z^3\end{align}$$
After finding $z$, for the final step we need to solve
$$\begin{cases}x+y=4-z\\ xy =f(z)\end{cases}$$
By the Vieta's formulas, we have
$$t^2-(4-z)t+f(z)=0$$
where, $$t_1=x,\, t_2=y \,\,\, \text{or}\,\,\,t_1=y,\, t_2=x $$
Thus for every $z$, we obtain the following solutions:
$$(z,x,y) \,\,\,\text{and}\,\,\, (z,y,x).$$
