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Maybe you could be more specific in terms of what kind of properties you're looking for, or why you're interested in these matrices in the first place. The (very restrictive!) properties which define these matrices surely generate more identities than one could possibly know what to do with.
It sounds like you are saying that every irreducible cubic has a splitting field which has Galois group $S_3$, which is false. Similarly, it is false that Q(\alpha) has trivial automorphism group. Or maybe I'm misreading?
+1, very interesting. I enjoy that neither the original asker nor the ultimate answerer were present on MO, and yet the connection was still made, and the answer found a home.
Ah, you're right. I misread the bolded text in 6.1 as an algorithm. While this isn't quite as explicit as I would have hoped, Alperin's "A Mathemtical Theory of Origami Constructions and Numbers" contains in Section 4 a proof that intersections of conics are origami-constructible. I'm not sure how hard it would be to do explicitly by following the proof. Perhaps the best solution is to email Alperin directly.
While I agree with the premise here, you can motivate multiplication mod n by performing $a$ tasks which each require $b$ hours...what time will it be at the end? Of course, this is really doing (integer) * (residue) and not (residue)*(residue), but then you have them observe that if you do your task $n$ times, $b$ is irrelevant, and remarkably, all that matters is how many times you perform the task mod n!!
(Actually, I agree with darij's comment, but the sort of rule you'd get for getting $y$ as a function of $x$ when $x$ and $y$ are only implicitly related is not something calculus students would think of as a rule).
For example, when you think of representations of $S_3$, say, do you think about the homomorphism of $S_3$ to $GL_2(\mathbb{R})$, or the various ways that permutation elements can move around points in $\mathbb{R}^2$?
@gowers: Interesting, I think of the "$f$" version as the more natural of the two. What's an action? You take a $g\in G$ and an $x\in X$, and you get a new $x'\in X$. That's precisely encoded by f. Taking in a $g\in G$ and outputting "a function which sends $x$'s to $x'$'s" seems to me to obfuscate the matter.