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Decide wether a given equation (over $\mathbb{Q}$) can be solved by radicals and, if it is possible, find a radical sequence to solve it.

The main problem is that I'm not sure what they want from me, because $x^6 + 2x^5 - 5x^4 - 5x^3 - 5x^2 + 2x + 1$ is a reciprocal palindromic equation, so it can be solved by substituting $x + \frac{1}{x} = y$, or $0 = x^6 + 2x^5 - 5x^4 - 5x^3 - 5x^2 + 2x + 1 = (x^2 + x + 1)(x^4 + x^3 - 7x^2 + x + 1)$ and since their degrees are less than $5$, they can be solved by radicals, hence I guess original equation can be solved by radicals. Is this the right approach? And what after that? How do I find its radical sequence and solve the equation by using that sequence?

Tito Piezas III
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Pan Miroslav
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    Observe that $$x^4 + x^3 - 7x^2 + x + 1=0$$ is also reciprocal. See also :
    https://math.stackexchange.com/questions/480102/quadratic-substitution-question-applying-substitution-p-x-frac1x-to-2x4x and https://math.stackexchange.com/questions/403025/equation-with-high-exponents     – lab bhattacharjee May 18 '17 at 08:20
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    "Is this the right approach ?" Absolutely. "How do I find its radical sequence " No problem : find a sequence for $x^2+x+1$, a sequence for $x^4+x^3-7x^2+x+1$, then combine the two. – Ewan Delanoy May 18 '17 at 08:22
  • I'm so happy that it should be solved that way. Anyway, for finding their sequence: I can find all its roots, create radical sequence by using them and that's it? Because I was confused by the sentece: "find a radical sequence and solve them" - I thought that it is actually somehow possible to firstly find a radical sequence and then use it to solve the equation. – Pan Miroslav May 18 '17 at 08:49
  • More generally, the reciprocal equation of $ax^2+bx+c = 0$ is $cx^2+bx+a = 0$. For the special case when it remains the same, it is better to call it a "palindromic equation". – Tito Piezas III Jan 18 '18 at 11:06

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