<s>[INST] Having $A_1=a+b+c$,$A_2=a^2+b^2+c^2$, $A_3=a^3+b^3+c^3$ - how to get $a,b,c$? Perhaps I'm just a bit dense at the moment - I've re-read some of my notes from monthes ago concerning elementary symmetric polynomials, and I find that I've no idea how to approach the "inverse" problem: if you have the -say- three values $A_0,A_1,A_2,A_3$ given, and you know 
$$ \begin{matrix} 
A_0  &=& a^0+b^0+c^0 &= 3 \\  
A_1  &=& a^1+b^1+c^1 &= 29 \\ 
A_2  &=& a^2+b^2+c^2 &= 315 \\ 
A_3  &=& a^3+b^3+c^3 &= 3653 \\ 
\end{matrix}$$ 
- just to give some example values, then
Q: how would I approach the finding of $a,b,c$ by some general path, which one could as well use for problems with more values/variables.
I guess,  that would involve somehow a matrix/an eigenvalue-formulation but I don't get an idea for the first step at the moment... 
(P.s.: I've no good idea for the tags; please feel free to improve my selection)
 [/INST] As $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca),$
$ A_1^2=A_2+2(ab+bc+ca)\implies ab+bc+ca=\frac{A_1^2-A_2}2$
Again, $a^3+b^3+c^3-3abc=(a+b+c)\{a^2+b^2+c^2-(ab+bc+ca)\}=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}$
$\implies A_3-3abc=A_1\left(A_1^2-3\frac{A_1^2-A_2}2\right)$
Express $abc$ in terms  $A_1,A_2,A_3$
So, the equation whose roots are $a,b,c$ is $x^3-A_1x^2+\frac{A_1^2-A_2}2x-abc=0$
</s>