Finding trinity of complex numbers 
I want to find all $x_1,x_2,x_3$ that satisfy these three equalities:
  $$x_1+x_2+x_3=6$$
  $$x_1^2+x_2^2+x_3^2=14$$
  $$x_1^3+x_2^3+x_3^3=36$$

So i don't know whether i should solve it using the same techniques i do with normal systems of equations or is there some tricks behind it?
So since I found this problem while doing algebraic problems I think there should be an algebraic solution to it. But I am really lost in this case, I don't knwow how to begin.
Any help would be appreciated.
 A: HINT:
Use $$(a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ca)$$
and $$a^3+b^3+c^3-3abc=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}$$
to find $ab+bc+ca, abc$
Then $a,b,c$ are the roots of $$t^3-(a+b+c)t^2+(ab+bc+ca)t-abc=0$$
A: The hard way:
The first equation is that of a plane with normal $(1,1,1)$ and this direction is that of an axis of symmetry of the problem.
We will rotate it to bring it to a $u$ axis, using the similarity transform
$$\begin{cases}x=\sqrt2 u-\sqrt3 v+w,\\y=\sqrt2u+\sqrt3v+w,\\z=\sqrt2u\ \ \ \ \ \ \ \ \ \ -2w.\end{cases}$$
Then
$$x+y+z=3\sqrt2u=6,\\
u=\sqrt2.$$
(A plane perpendicular to $u$.)
$$x^2+y^2+z^2=6(u^2+v^2+w^2)=14,\\v^2+w^2=\frac13.$$
(A cylindre of axis $u$.)
$$x^3+y^2+z^3=18\sqrt2uv^2+18\sqrt2uw^2+6\sqrt2u^3+18wv^2-6w^3\\
=18\sqrt2uv^2+18\sqrt2uw^2+18\sqrt2uu^2-12\sqrt2u^3+18wv^2-6w^3\\
=42\sqrt2u-12\sqrt2u^3+18wv^2-6w^3=36,\\
3wv^2=w^3.
$$
(Three planes intersecting on $u$.)
This gives the six solutions
$$v=\pm\frac1{\sqrt3},w=0\lor v=\pm\frac1{2\sqrt3},w=\pm\frac12,$$
i.e.
$$\begin{cases}x=2\mp1,\\
y=2\pm1,\\
z=2,\end{cases}$$
or
$$\begin{cases}x=2\mp\dfrac12\pm\dfrac12,\\
y=2\pm\dfrac12\pm\dfrac12,\\
z=2\ \ \ \ \ \ \ \ \ \mp1.\end{cases}$$
As expected, the six permutations of $(1,2,3)$.
