Symmetric polynomials and g non symmetric If $g=x_1+2x_2+3x_3, s_1=x_1+x_2+x_2, s_2=x_1x_2+x_1x_3+x_2x_3$ and $s_3=x_1x_2x_3$ , write $x_1, x_2$ and $x_3$ in function of $g, s_1, s_2$ and $s_3$.
 A: I'm pretty sure a complete answer is too long for this format, but I will give a sketch of my approach.
First off, while we could try to find relations using powers of $g$, the bookkeeping will be awful.  To try and ameliorate this, instead consider
$$
(g-2\sigma_1)^2 = (x_3-x_1)^2 = x_3^2+x_1^2 - 2x_1x_3
$$
This element is fixed by the Galois action interchanging $x_1$ and $x_3$.  It follows that this element generates the fixed points of this action, and thus we can in theory write $x_2$ in terms of this element.  The minimal polynomial for this element is degree $3$ over the base field $\Bbb Q(\sigma_1, \sigma_2, \sigma_3)$, so there are symmetric polynomials $F_1, F_2, F_3, F_4$ such that
$$
x_2 = \frac{F_1 + (g-2\sigma_1)^2F_2 + (g-2\sigma_1)^4F_3}{F_4}
$$
If we rewrite the desired relation as 
$$
 F_1 + (g-2\sigma_1)^2F_2 + (g-2\sigma_1)^4F_3- x_2F_2 = 0
$$
we see that we might as well assume that the $F_i$ are homogeneous.  Some counting lets us see that we must have a solution in total degree $4$, since we can take arbitrary linear combinations of the following $10$ elements:
$$
\begin{array}{lll}
(g-2\sigma_1)^4 & 
(g-2\sigma_1)^2\sigma_2 & 
(g-2\sigma_1)^2\sigma_1^2 \\
x_2\sigma_3 &
x_2\sigma_2\sigma_1 & 
x_2\sigma_1^3 \\
\sigma_3\sigma_1 &
\sigma_2^2 &
\sigma_2\sigma_1^2\\
\sigma_1^4,
\end{array}
$$
and the space of degree-4 polynomials symmetric in $x_1$ and $x_3$ is $9$-dimensional, spanned by
$$
\begin{array}{llll}
x_2^4 &
x_1^4+x_3^4 & 
x_2^3(x_1+x_3) \\
x_2(x_1^3+x_3^3) &
x_1x_3(x_1^2+x_3^2) & 
x_2^2(x_1^2+x_3^2) \\
x_1^3x_3^3 &  
x_2^2x_1x_3 &
x_2x_1x_3(x_1+x_3)
\end{array}
$$
At this point, finding suitable $F_i$ is "just" linear algebra.  After obtaining an expression for $x_2$, you can get $x_3 = \frac{g- \sigma_1 - x_2 }{2}$, $x_1 = g-2x_2-3x_3$.
