I'm dealing with the expression $x = \frac{1}{3}y(y+1)(2y+1)^2(2y^2+2y+1)$. What is this approximately, if one is explicitly writing y in terms of x? There's no general formula for sixth powers unfortunately.


Also can one given an approximation of this so that the difference between the true y and the approximation go to zero? (Not just the ratio).



Solution to sixth order equation
For $x,y>0$ there is a unique solution $y(x)$ to $x = \frac{1}{3}y(y+1)(2y+1)^2(2y^2+2y+1)$ given by
$$y=\tfrac{1}{2} 3^{-1/3} \sqrt{\frac{\left(\sqrt{11664 x^2-3}+108 x\right)^{2/3}+3^{1/3}}{\bigl(\sqrt{11664 x^2-3}+108 x\bigr)^{1/3}}}- \tfrac{1}{2}.$$
Here is a plot of $y$ versus $x$.


![](https://i.stack.imgur.com/b483J.png)
