## Solvable sextics

In a previous post, we discussed the Bring-Jerrard quintic,

$x^5+ax+b=0$

and rational parametrizations for {a, b}.  It turns out this has a sextic version. To illustrate, these irreducible equations are solvable,

$\text{(1)}\;\; x^6+3x+3 = 0$

$\text{(2)}\;\; x^6-7^2x-7^2 = 0$

However, each needs a different approach. The first factors into two cubics over $\sqrt{-3}$, 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:

$x^6+4mn(m^2-3n)x+n(m^4+18m^2n+n^2) = 0$

This factors over $\sqrt{-n}$.  Let {m, n} = {-1/2, 3/4}, and it will yield (1).

(2)  2nd family:

The second is trickier.  Given,

$x^2+vx+(pv^2-qv-nr) = 0$

$v^3-rv-r = 0$

Eliminating v between them using resultants (easily done in Mathematica or Maple), results in a sextic in the variable x of form,

$x^6+ax+b = 0$

if,

\begin{aligned}p &=\frac{3n+1}{2}\\q &=\frac{9n+1}{4}\\r &= \frac{(9n+1)(9n-7)}{4(3n+1)(n-1)} \end{aligned}

Let n = 3, we get {p, q, r} = {5, 7, 7}, and it yields (2).  Thus, to solve,

$x^6-7^2x-7^2 = 0$

entails solving the cubic in v,

$v^3-7v-7 = 0$

though other n do not show this nice symmetry.  Another good value is at n = 1/3 which gives,

$x^6+3x+5 = 0$

### 2 responses to this post.

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