The following three cute irreducible octics are solvable,

However, each needs to be solved in a different way: they need a quadratic, quartic, and septic subfield, respectively.

The first is the easiest, it factors over into two quartics. The second does not factor over a square root extension, but factors into four quadratics,

where the coefficients are determined by the quartic,

The command,

Resultant[x^2+vx -(2v^3-7v^2+5v+33)/13 , v^4+22v+22 ,v]

done in *Mathematica* or in *www.wolframalpha.com* will eliminate the variable *v* and recover [2]. The first two octics are by this author. (*Any other example for the second kind with small coefficients?*)

The third (by Igor Schein) is the hardest, as it needs a septic subfield. Interestingly though, the solution involves the 29th root of unity. Given,

Then,

where the 7 constants {a, b, c, d, e, f, g} is the *square root* of the *appropriate* root of the solvable septic,

namely,

Note that,

hence one can use the roots of this octic to express the roots of its resolvent septic, and vice versa. In terms of radicals, Peter Montgomery expressed the septic roots as,

where *w* is *any* complex root of unity (excluding *w* = 1) such that . For example, will yield the value for , and so on.