Geometric distinction between real cubics with different Galois group? The following Cubics have 3 real roots but the first has Galois group $C_3$ and the second $S_3$


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*$x^3 - 3x + 1$ (red)

*$x^3 - 4x + 2$ (green)


Is there any geometric way to distinguish between the two cases? Obviously graphing this onto the real line does not help.

It is not clear to me why you cannot transpose the red dots but you can transpose the green ones.
 A: There's no reason to expect that the set of real points tells you the full story in an arithmetic situation. For example, can you tell that $\pi$ is transcendental but that $\sqrt{10}$ isn't from looking at their relative positions on the number line? 
One thing you can do which (depending on your tastes) could count as geometric is looking at Frobenius elements.
Proposition: Let $f(x) \in \mathbb{Z}[x]$ be a monic irreducible polynomial. Suppose that for some prime $p$ not dividing the discriminant, $f$ splits into irreducible factors of degrees $d_1, d_2, ... d_k$. Then there is an element of cycle type $(d_1, d_2, ... d_k)$ in the Galois group of $f$.
Hence you can find a transposition in the Galois group (and show that the Galois group is $S_3$) by finding a prime $p$ relative to which $f$ splits as the product of a linear and an irreducible quadratic factor. (This is completely analogous to the situation over $\mathbb{R}$: one might say that complex conjugation is the "Frobenius element at infinity.")
Geometrically, instead of looking at the real points of the scheme $\text{Spec } \mathbb{Z}[x]/f(x)$, we look at points over finite fields. But from a scheme-theoretic point of view all of these points are included in the geometry of the "arithmetic curve" $\text{Spec } \mathbb{Z}[x]/f(x)$. 
The Frobenius density theorem even guarantees that the converse holds: for every cycle type in the Galois group there is a prime $p$ realizing that cycle type.
A: Almost all cubics (with integer coefficients and three real roots) have Galois group $S_3$. What exactly is meant by "almost all" is a little technical, but the phrase can be made precise, and the result rigorously proved. One consequence is that if you start with a $C_3$ cubic and perturb the roots the tiniest little bit then with probability $1$ you now have an $S_3$ cubic. So just looking at the red dots can't help you: it's guaranteed that there is a set of green dots so close by that you wouldn't be able to distinguish them with an electron microscope. 
