How do you find the roots of a degree $11$ polynomial (or any other prime degree) using langrange resolvents? In particular, what transformation gets you from the given polynomial to the auxiliary equation? A simple example is a generalization of the Moivre quintic $$x^{11}+11bx^{9}+44b^{2}x^{7}+66b^{3}x^{5}+44b^{4}x^{3}+11b^{5}x+c=0$$ for which the auxillary polynomial is $$y^{2}+cy-b^{11}=0$$ How would you get $y^{2}+cy-b^{11}=0$ from $x^{11}+11bx^{9}+44b^{2}x^{7}+66b^{3}x^{5}+44b^{4}x^{3}+11b^{5}x+c=0$? This example doesn't actually require Langrange resolvents, since generalized examples exist for nonprime degrees, like the degree $9$ polynomial $$x^{9}+9bx^{7}+27b^{2}x^{5}+27b^{3}x^{3}+9b^{4}x+c=0$$ or more generally, $$x^{n}+nbx\sqrt{\left(x^{2}+b\right)^{n-3}}+2c=0,\ x=\sqrt[n]{\sqrt{b^{n}+c^{2}}-c}-\sqrt[n]{\sqrt{b^{n}+c^{2}}+c}$$ but I don't know how other examples would be solved. I'm asking about the degree $11$ case because I know other methods to solve quintics and septics.
EDIT: Parametrized special case polynomials with formulas written directly in the answer would be preferable, so formatted as $$P\left(x\right)=0,\ x=f\left(a,\ b,\ c,\ d,\ ...\right)$$ along with an explanation of where it comes from.