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,