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 $x,y,z$
 A: 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}
A: 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:
$$(z,x,y) \,\,\,\text{and}\,\,\, (z,y,x).$$
A: $x+y+z=4$
$x^2+y^2+z^2=14$
$xy+yz+zx=1$(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
$(3,2,-1)$.
