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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).

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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 $x$.

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    $\begingroup$ To give an idea of how you can come up with this, you can realize that substituting $z=y+\frac{1}{2}$, you obtain $x$ as an even function of $z$ (a way to suspect this is noticing that the roots of the given degree 6 polynomial are symmetric respect to $-\frac{1}{2}$), so you can call another variable $t=z^2$ and then you have $x$ as a cubic polynomial in $t$, which allows you to use Cardano´s formula for the cubic equation. $\endgroup$
    – Saúl RM
    Commented Dec 12, 2021 at 18:58
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    $\begingroup$ Since the RHS is strictly increasing, zero at zero and tends to infinity, existence and uniqueness of a positive solution is evident. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 13, 2021 at 1:12
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    $\begingroup$ Note that $108^{2}=11664$. By introducing $\xi:=108 x$ the solution's formula becomes rather compact. $\endgroup$
    – Johannes Trost
    Commented Dec 13, 2021 at 11:08

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