In a previous post, we discussed the Bring-Jerrard quintic,
and rational parametrizations for {a, b}. It turns out this has a sextic version. To illustrate, these irreducible equations are solvable,
However, each needs a different approach. The first factors into two cubics over , while the second factors into three quadratics whose coefficients are determined by a cubic. These two examples belong to two infinite (but not complete) families.
(1) 1st family:
This factors over . Let {m, n} = {-1/2, 3/4}, and it will yield (1).
(2) 2nd family:
The second is trickier. Given,
Eliminating v between them using resultants (easily done in Mathematica or Maple), results in a sextic in the variable x of form,
if,
Let n = 3, we get {p, q, r} = {5, 7, 7}, and it yields (2). Thus, to solve,
entails solving the cubic in v,
though other n do not show this nice symmetry. Another good value is at n = 1/3 which gives,