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 $a,b,c$?

Question

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)

Answer 1

Hint:

Find the $ab+ac+bc$ and $abc$ .

The roots of the polynomial $P(x)=x^3-(a+b+c)x^2+(ab+bc+ac)x-abc$ are $a,b,c$.

Answer 2

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 $x^3-A_1x^2+\frac{A_1^2-A_2}2x-abc=0$