Deradicalization of denominators

Question

Task: Develop a fraction equivalent to $$ 1\over{\sum\limits_{i=0}^{n-1}c_in^{i/n}} $$ in which the denominator is rational.

Answer 1

Step 1: Develop the denominator. The objective will be a fraction in which the numerator and denominator are two nearly-similar determinants which differ only in their top rows. The initial form of the denominator is

$\begin{vmatrix} c_0&c_{n-1}n&c_{n-2}n&\cdots&c_3n&c_2n&c_1n\\ c_1&c_0&c_{n-1}n&\cdots&c_4n&c_3n&c_2n\\ c_2&c_1&c_0&\cdots&c_5n&c_4n&c_3n\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ c_{n-3}&c_{n-4}&c_{n-5}&\cdots&c_0&c_{n-1}n&c_{n-2}n\\ c_{n-2}&c_{n-3}&c_{n-4}&\cdots&c_1&c_0&c_{n-1}n\\ c_{n-1}&c_{n-2}&c_{n-3}&\cdots&c_2&c_1&c_0\\ \end{vmatrix}$.

Step 2: The numerator is merely the same determinant as the denominator, except that its top row is replaced by

$\begin{matrix} 1&\sqrt[n]n&\sqrt[n]{n^2}&\cdots&\sqrt[n]{n^{n-3}}&\sqrt[n]{n^{n-2}}&\sqrt[n]{n^{n-1}}\\ \end{matrix}$.

This gives rise, for example, to the following:

${1\over{A+B\sqrt n}}= {{\begin{vmatrix} 1&\sqrt n\\ B&A\\ \end{vmatrix}}\over {\begin{vmatrix} A&Bn\\ B&A\\ \end{vmatrix}}}={{A-B\sqrt n}\over{A^2-B^2 n}}$ and
${1\over{A+B\sqrt[3] n+C\sqrt[3]{n^2}}}= {{\begin{vmatrix} 1&\sqrt[3]n&\sqrt[3]{n^2}\\ B&A&Cn\\ C&B&A\\ \end{vmatrix}}\over{ \begin{vmatrix} A&Cn&Bn\\ B&A&Cn\\ C&B&A\\ \end{vmatrix}}}={{A^2-BCn+(C^2n-AB)\sqrt[3]n+(B^2-AC)\sqrt[3]{n^2}}\over{A^3+B^3 n+C^3 n^2-3ABCn}}$.

An example of how a typical denominator (between the bars) is set up:

$\begin{array}{ccccc|cccccc|ccccc} F&E&D&C&B&A&Fn&En&Dn&Cn&Bn&&&&&\\ &F&E&D&C&B&A&Fn&En&Dn&Cn&Bn&&&&\\ &&F&E&D&C&B&A&Fn&En&Dn&Cn&Bn&&&\\ &&&F&E&D&C&B&A&Fn&En&Dn&Cn&Bn&&\\ &&&&F&E&D&C&B&A&Fn&En&Dn&Cn&Bn&\\ &&&&&F&E&D&C&B&A&Fn&En&Dn&Cn&Bn\\ \end{array}$