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As the Galois theory, all solution of a polynomial with a solvable Galois group cannot be expressed by form of radicals. This mean, even for unsolvable polynomials, it is possible that some of the solutions can be written in the form of radicals. Can we find one such unsolvable polynomial? Which is its root solution.

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    $\begingroup$ Examples can be found in this related question (which appeared in the links on the right side of the page for this question). $\endgroup$ Commented Jun 14, 2022 at 7:16
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    $\begingroup$ For irreducible polynomials, all roots are conjugate under Galois so one of the roots can be expressed with radicals iff every root can. $\endgroup$
    – Aphelli
    Commented Jun 14, 2022 at 7:22
  • $\begingroup$ @GregMartin I realize that your comments are contradictory to Aphell's. and did not find the corresponding polynomial in your link $\endgroup$
    – mayi
    Commented Jun 14, 2022 at 7:44
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    $\begingroup$ Greg Martin's comment is about the question in the title, only meant to address the requirement of an unsolvable Galois group. If you want a polynomial with an unsolvable Galois group with some roots still in terms of radicals, you must use a reducible polynomial like $$(x^2-2)(x^5+2x+1).$$ $\endgroup$ Commented Jun 14, 2022 at 12:43
  • $\begingroup$ @JyrkiLahtonen What you gave is indeed a reducible polynomial, a reducible polynomial is not possible to satisfy my requirement, right? Do you have any basis (theorem)? $\endgroup$
    – mayi
    Commented Jun 14, 2022 at 16:49

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