Solve this sextic I'm working with the expression $m = 32an^6+96an^5+120an^4+80an^3+28an^2+4bn^2+4an+4bn+2a+2c
$. What exactly is the closed radical form of this, if one were to write $n$ in terms of $m$? There's no general formula for sixth powers but I want to know if this particular class can be given in radicals for all $a, b \in \mathbb{R}$.

Mathematica finds a closed-form expression for the solutions $n$ as a function of $m$ of the equation
$$m = 32an^6+96an^5+120an^4+80an^3+28an^2+4bn^2+4an+4bn+2a+2c.$$
The expressions for general $a,b,c$ are lengthy. By way of example, for $a=1$, $b=2$, $c=-3$ the two real solutions are
$$n=-\tfrac{1}{2}\pm\tfrac{1}{2}\sqrt{f_m^{1/3}-f_m^{-1/3}},\;\;f_m=\sqrt{m^2+12 m+37}+m+6.$$

More generally, two of the solutions are
$$n=-\frac{1}{2}\pm\frac{1}{8}\sqrt{\frac{2^{2/3} \left(\sqrt{4 s^3+z^2}+z\right)^{2/3}- 2^{4/3} s}{3a \left(\sqrt{4 s^3+z^2}+z\right)^{1/3}}},$$
$$z=27648 a^2 (b-2a)+27648 a^2 (m-2c),\;\;s=192 a(2b-a).$$
I have not checked the parameter range where these two solutions are real.

