Solvable sextics

10 Apr

Solvable sextics

Posted April 10, 2012 by tpiezas in algebra, equations, mathematics. Tagged: cubics, quadratic, sextics. 2 Comments

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.

Posted by paramanands on April 13, 2012 at 4:37 am

Hello Tito,
Nice to have you on wordpress. I loved your content in the google pages and I hope to see the similar exciting content here.

Paramanand

Reply
- Posted by tpiezas on April 13, 2012 at 1:48 pm
  
  Dear Paramanand,
  
  I have you to thank for. I didn’t know WordPress supported LaTex. I tried Blogger before but there was no efficient way to display equations. If I had known about WordPress, I would have done this years ago! So thanks again. 🙂
  
  Reply

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