# 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 $$a, b \in \mathbb{R}$$.

• Erm, when is that sextic function one-to-one? Jun 23 at 4:23

The RHS only depends on $$n(n+1)$$, specifically it can be written as $$4a\left(8(n(n+1))^3+6(n(n+1))^2+n(n+1)+\frac12\right)+4bn(n+1)+2c,$$ so you have to use the standard formulas to solve a cubic equation $$g(n(n+1))=m$$, and then a quadratic equation $$n(n+1)=k$$, which can be done in radicals.
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.