# 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:

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

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}
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.
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:$$\alpha_1^2+\alpha_2^2+\alpha_3^2=(\alpha_1+\alpha_2+\alpha_3)^2-2(\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_3\alpha_1)=a^2-2b$$