Suppose we want to solve$$x^3 - ax^2 + bx - c = 0.$$We know a priori that this can be factored as $(x - r_0)(x - r_1)(x - r_2)$; by Vieta's formulas, we know$$a = r_0 + r_1 + r_2,\quad b = r_0r_1 + r_1 r_2 + r_2r_0,\quad c = r_0r_1r_2.$$These expressions are invariant under three-cycles. Now, we make the substitution$$r_0 = u_0 + u_1 + u_2,\quad r_1 = u_0 + u_1\omega + u_2\omega^2,\quad r_2 = u_0 + u_1\omega^2 + u_2\omega^4,$$where $\omega$ is a primitive cube root of unity. Explicitly, we have$$u_0 = {1\over3}(r_0 + r_1 + r_2),\quad u_1 = {1\over3}(r_0 + r_1\omega^{-1} + r_2 \omega^{-2}),\quad u_2 = {1\over3}(r_0 + r_1 \omega^{-2} + r_2\omega^{-4}).$$Conceptually,$$\text{if }F(z) = u_0 + u_1z + u_2z^2,\text{ then }r_i = F(\omega^i).$$The first of Vieta's relations now reads$$a = F(1) + F(\omega) F(\omega^2) = 3u_0,$$which is a roots of unity filter on $F$. The second reads$$b = F(1)F(\omega) + F(\omega)F(\omega^2) + F(\omega^2)F(1),$$which is a roots of unity filter on$$F(z)F(\omega z) = (u_0 + u_1 z + u_2z^2)(u_0 + u_1z\omega + u_2z^2\omega^2).$$Since in a filter, we only care abou tthe cubic terms – here $u_0^2$ and $-u_1u_2z^3$ – we find that $b = 3u_0^2 - 3u_1u_2$. If we repeat the same thing for $c$ and compile everything together, we arrive at$$a = 3u_0,\quad b = 3u_0^2 - 3u_1u_2,\quad c = u_0^3 + u_1^3 + u_2^3 - 3u_0u_1u_2.$$The first equation gives us $u_0 = a/3$ for free, and we can then obtain $u_1u_2$ in terms of $a$, $b$, $c$. Finally, the last equation tells us what $u_1^3 + u_2^3$ are. So, we can compute the values of $u_1^3 + u_2^3$, $u_1^3 \cdot u_2^3$; this reduces to a quadratic, which we can solve.

My question is, what is the precise underlying interplay going on between the Galois theory and the "finite" Fourier analysis here, if there is any? Am I shooting in the dark for something that may possibly not exist? Or is there something deep behind all this?