polynomial equation $x^6+2x^5-5x^4+9x^3-5x^2+2x+1=0$ in terms of radicals

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The task is to solve $$x^6+2x^5-5x^4+9x^3-5x^2+2x+1=0$$ by radicals. A hint is given to put $$u=t+1/t.$$ I tried factoring the polynomial and applying the rational root theorem, none gave me direction. I also don't know how to apply the hint.

Any suggestion or help will be very much appreciated. Thank you.

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Hint: the polynomial is self-reciprocal (or palindromic) and $x=0$ is obviously not a root, so divide by $x^3$ and regroup to get:

$$x^3 + \frac{1}{x^3} +2\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)+9=0$$

Let $u = x + \frac{1}{x}$ so that $x^2 + \frac{1}{x^2} = u^2 - 2$ and $x^3+\frac{1}{x^3} = u^3 - 3u$ then the equation reduces to the cubic in $u$:

$$u^3+2 u^2 - 8 u + 5 = 0$$

which factors as:

$$(u-1)(u^2+3u-5)= 0$$

Each of the $3$ roots in $u$ gives a quadratic in $x$ to solve for the total of $6$ roots in $x$.