How do I solve this equation for R/r?

Question

How do I solve this expression for $\frac{R}{r}$?

$\frac{1}{(r-R)^3}-\frac{y}{R^2(r-R)}=\frac{1}{r^3}$

Answer 1

Rearrange the equation to get

$$1 - y \left ( 1 - \frac{r}{R} \right )^2 = \left ( 1 - \frac{r}{R} \right )^3 $$

Set $z = 1-r/R$ and get the equation:

$z^3+y z^2-1=0$

Get a root at $z_0(y)$ numerically or otherwise. Then $R/r$ is then

$$\frac{R}{r} = \frac{1}{1-z_0(y)} $$

NB There is an analytical expression for $z_0$:

$$ z_0(y) = \frac{1}{3} \left(\frac{\sqrt[3]{-2 y^3+3 \sqrt{3} \sqrt{27-4 y^3}+27}}{\sqrt[3]{2}}+\frac{\sqrt[3]{2} y^2}{\sqrt[3]{-2 y^3+3 \sqrt{3} \sqrt{27-4 y^3}+27}}-y\right) $$