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I have the equation $x^6 + 3 x^5 + 3 x^4 + 3 x^3 + 2 x^2 + 1 = 0$ and I want to find the roots in terms of radicals, if possible.

According to Wikipedia, "It follows from Galois theory that a sextic equation is solvable in term of radicals if and only if its Galois group is contained either in the group of order 48 which stabilizes a partition of the set of the roots into three subsets of two roots or in the group of order 72 which stabilizes a partition of the set of the roots into two subsets of three roots."

I've tried to figure out what Galois group my equation is in, without success. Galois theory is way over my head.

Is this equation solvable or not?

Lawton
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    the Galois group is the full symmetric group on six letters, order 720 – Will Jagy Jan 28 '21 at 15:36
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    The polynomial is irreducible over $\Bbb Q$ and its Galois group is not solvable. So you cannot find a formula in terms of radicals. But there are other methods to find, say, real roots. They are given by $x= - 1.94079980653$ and $x= - 1.31070264134$. – Dietrich Burde Jan 28 '21 at 15:46

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You can compute Galois groups of polynomials using the open-source computer algebra system Sage. The documentation for Sage's tools for working with Galois groups can be found here. To give an example:

A.<x> = QQ[]   # defines A = Q[x] to be the polynomial ring in x over the rationals
poly = x^6 + 3*x^5 + 3*x^4 + 3*x^3 + 2*x^2 + 1
G = poly.galois_group()
G.is_solvable()

If you want to do the computation by hand (or understand how computer implementations work), the book A Course in Computational Number Theory by Cohen (Graduate Texts in Mathematics 138, Springer-Verlag, 1993) is a good reference for efficient algorithms. In particular, section 6.3.5 deals with the case of determining the Galois group of a sextic polynomial. The method described there uses resolvent polynomials to distinguish between the 16 possible transitive subgroups of $S_6$.

Daniel Hast
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