Finding product and power combinations of roots of cubic Let $\alpha_1,\alpha_2,\alpha_3 \in \mathbb{C}$ be the roots of the polynomial $f(x)=x^3-x+z$. What is the irreducible polynomial of $\beta=\alpha_1^k\alpha_2^j$ over $\mathbb{Q}$ for positive integers k and j?
having either j,k=0 is trivial
j,k=1 gives
$\;x^3-x^2-1$.
j=1,k=2 results in the following sextic, which I beleive can be rededuced to two cubics, but I don't know how:
$x^6-3zx^5+(6z^2-1)x^4+(2z-7z^3)x^3+(6z^4-z^2)x^2-3z^5x+z^6=0$
Edit:
wolframalpha  seems to suggest the sextic is solvable (in this case , letting z =1/5)
https://www.wolframalpha.com/input/?i=x%5E6-3*%281%2F5%29x%5E5%2B%286%281%2F5%29%5E2-1%29x%5E4%2B%282*%281%2F5%29-7*%281%2F5%29%5E3%29x%5E3%2B%286*%281%2F5%29%5E4-%281%2F5%29%5E2%29x%5E2-3*%281%2F5%29%5E5*x%2B%281%2F5%29%5E6%3D0
the output suggests the sexic is of the form $(a-x)^6+b(a-x)^4+c(a-x)^2+d$
equating the coefficients a,b,c,d gives the desired result
$a=z/2,b=9z^2/4-1,c=(z^2/16)(27z^2+8),d=(z^4/64)(27z^2-4)$
this is obviously solvable
remains to be de determined if this works for all j,k
 A: $p_j^{(f)}=\alpha_1^j+\alpha_2^j+\alpha_3^j$ can be calculated in terms of the coefficients of $f(x)$ by Newton's identities.
The polynomial $g(x)=x^3f\left(\frac{1}{x}\right)=z\,x^3-x^2+1$ has roots $\,\frac{1}{\alpha_p}\,$, so $p_k^{(g)}=\frac{1}{\alpha_1^k}+\frac{1}{\alpha_2^k}+\frac{1}{\alpha_3^k}$ can also be calculated in terms of $z$ using the same identities.
For $j=k$ it follows that $\beta_1=\alpha_2^k\alpha_3^k, \beta_2=\alpha_1^k\alpha_3^k, \beta_3=\alpha_1^k\alpha_2^k$ are the roots of the cubic $t^3 - e_1 t^2 + e_2 t - e_3$ where:

*

*$e_1=\beta_1+\beta_2+\beta_3=\left(\alpha_1\alpha_2\alpha_3\right)^k \left(\frac{1}{\alpha_1^k}+\frac{1}{\alpha_2^k}+\frac{1}{\alpha_3^k}\right)=(-z)^k \,p_k^{(g)}$


*$e_2 = \beta_1\beta_2+\beta_1\beta_3+\beta_2\beta_3=\left(\alpha_1\alpha_2\alpha_3\right)^k \left(\alpha_1^k+\alpha_2^k+\alpha_3^k\right)=(-z)^k\,p_k^{(f)}$


*$e_3 = \beta_2\beta_2\beta_3=\left(\alpha_1\alpha_2\alpha_3\right)^{2k}=z^{2k}$
For $j \ne k\,$, it follows using a similar argument that $x=\alpha_i^k$ will be the roots of a cubic $k(x)$, and $y=\alpha_i^j$ the roots of a different cubic $j(y)$. The products $\beta=\alpha_i^k\alpha_{i^i}^j \,\big|_{i \ne i'}$ will be the roots of a sextic, which is irreducible in the general case.
Computationally, the polynomial in $t=xy$ can be determined by eliminating $x$ between $k(x)=0$ and $j(t/x)=0$ using for example resultants. This will give a polynomial of degree $3 \cdot 3=9$ in $t$, but it is known that three of the roots $\alpha_1^{k+j}, \alpha_2^{k+j}, \alpha_3^{k+j}$ satisfy a cubic which can be factored out, leaving a polynomial of degree $9 - 3 = 6\,$.
