Solving for coefficients of a polynomial in terms of roots of another This was a homework assignment a while back (already turned in) that I'm not totally sure how to approach. I think it might be simple but I can't wrap my mind around it.
If someone could give me a tip that would be great.
I'm given a polynomial $f=x^3+ax^2+bx+c$ with roots $\alpha_1,\alpha_2,\alpha_3$.
How do I find the coefficients of the following two polynomials, in terms of a,b, and c:


*

*the monic cubic with roots $\alpha_1^2,\alpha_2^2,\alpha_3^2$

*the monic cubic with roots $\alpha_1+\alpha_2,\alpha_1+\alpha_3,\alpha_2+\alpha_3$


What I know:
I understand that by viete's formula, $a = \alpha_1+\alpha_2+\alpha_3$, $b=\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3$, and $c=\alpha_1\alpha_2\alpha_3$. 
But is there an elegant way to solve for the relationship between the coefficients for (1) and (2) using the coefficients of $f$? It seems resistant to a clean solution, since the relationship isn't necessarily linear.
Cheers.
 A: Hint: $\alpha_1^2+\alpha_2^2+\alpha_3^2=(\alpha_1+\alpha_2+\alpha_3)^2-2(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3)=a^2-2b$; there's one of the coefficients for the first polynomial you want. By the way, your formulas for the coefficients are missing some minus signs.
A: There is a general method to express symmetric poynomials in terms of the elementary symmetric polynomials. 
For example for the cubic with roots $\alpha_1^2, \alpha_2^2,\alpha_3^2$ on needs to express $\alpha_1^2+\alpha_2^2+\alpha_3^2$, $\alpha_1^2\alpha_2^2+\alpha_2^2\alpha_3^2+\alpha_1^2\alpha_3^2$, and $\alpha_1^2\alpha_2^2\alpha_3^2$ in terms of $\alpha_1+\alpha_2+\alpha_3$, $\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3$, and $\alpha_1\alpha_2\alpha_3$. One can do so by eliminating from highest to lowest power of the polynomial with the most distinct factors. For example,
$$\begin{align}\alpha_1^2+\alpha_2^2+\alpha_3^2&=(\alpha_1+\alpha_2+\alpha_3)^2-2\alpha_1\alpha_2-2\alpha_1\alpha_3-2\alpha_2\alpha_3\\
&=a^2-2b\end{align} $$
$$\begin{align}\alpha_1^2\alpha_2^2+\alpha_2^2\alpha_3^2+\alpha_1^2\alpha_3^2&=(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3)^2-2\alpha_1^2\alpha_2\alpha_3-2\alpha_1\alpha_2^2\alpha_3-2\alpha_1\alpha_2\alpha_3^2\\
&=b^2-2\alpha_1\alpha_2\alpha_3(\alpha_1+\alpha_2+\alpha_3)\\&=b^2-2ca\end{align} $$
A: You could try rearrangements - e.g. for the first one:$$(\alpha_1+\alpha_2+\alpha_3)^2=\alpha_1^2+\alpha_2^2+\alpha_3^2+2(\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_3\alpha_1)$$Hence:$$\alpha_1^2+\alpha_2^2+\alpha_3^2=(\alpha_1+\alpha_2+\alpha_3)^2-2(\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_3\alpha_1)=a^2-2b$$
Then follow a similar route for the rest.
A: You did not use keyword "elementary symmetric polynomials", which are expressions for a, b, c (with proper signs), so I assume you missed on the importance of the notion. It turns out that every symmetric polynomial can be expressed as an algebraic combination of elementary symmetric polynomials. The theorem proof is constructive, that is, it demonstrates an algorithm to do so. Not sure what you mean by "clean solution", in practice it is never a problem to express given symmetric polynomial as a combination of elementary following either an algorithm from a theorem, or using ad hoc properties of specific problem.
