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Can all the roots (real or complex) of $x^6 + x^5 +x^4 + x^3 +x^2 + x = n$ be found using trigonometric methods? Many thanks to all of answers.

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  • $\begingroup$ In general probably not, since I see no reason why the corresponding Galois group should be solvable. $\endgroup$ – Potato Jun 18 '14 at 18:31
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$$x^6+x^5+x^4+x^3+x^2+x=n\\x^6+x^5+x^4+x^3+x^2+x+1=n+1\\(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=(n+1)(x-1)\\x^7-1=(n+1)(x-1)\\ x=r*e^{ia}\\r^7*e^{7ia}-1=(n+1)(r*e^{ia}-1) $$

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