Rewriting to quintic equation

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Consider the force $F_\Omega$: $$ F_\Omega= \Omega^2\left(x -\frac{\beta\left(x+\alpha R\right)R^3}{\left(\left(x+\alpha R\right)^2\right)^{3/2}} -\frac{\alpha\left(x-\beta R\right)R^3}{\left(\left(x-\beta R\right)^2\right)^{3/2}} \right) $$ Where $\alpha+\beta=1$

(also to the attentive reader; the unit of mass, which is missing, has been set equal to unity)

Set $F_\Omega=0$

Now, make the substitution $x=R(u+\beta)$

The following three quintics can be obtained $$ u^2\left(\left(1-s_1\right)+3u+3u^2+u^3\right) =\alpha\left(s_0+2s_0u+\left(1+s_0-s_1\right)u^2+2u^3+u^4\right) $$ Here $s_0=\mbox{sign}(u)$ and $s_1=\mbox{sign}(u+1)$

Now to my question: How can we arrive at the quintic equation?

The equations come from a paper on Lagrange-points: http://www.physics.montana.edu/faculty/cornish/lagrange.pdf

It has been suggested to me (from the author) that I should

1) Make the substitution of $x$

2) Put everything on a common denominator

3) Make the numerator vanish by taking the term to the $\frac{3}{2}$-power to the left side of the equation

I have tried of course done the substitution, but fail to put it all on a common demonitor, and honestly can't see for myself where all these $u$'s come into the picture.

Feel free to put on some relevant tags. I couldn't think of any.

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Presumably the variable $R$ is understood to be positive. Assuming this, let's see what the substitution $x=uR+\beta R$, with $\alpha+\beta=1$, does to the three pieces of the expression inside the brackets:

$${\alpha(x-\beta R)R^3\over((x-\beta R)^2)^{3/2}}={\alpha uR^4\over|uR|^3}={\alpha Rs_0\over u^2}$$

where $s_0$ is the sign of $uR$, which is presumably the same as the sign of $u$.

$${\beta(x+\alpha R)R^3\over((x+\alpha R)^2)^{3/2}}={\beta(u+\beta+\alpha)R^4\over|(u+\beta+\alpha)R|^3}={(1-\alpha)Rs_1\over(u+1)^2}$$

where $s_1$ is the sign of $(u+1)R$, which is presumably the same as the sign of $u+1$.

$$x=uR+\beta R=(u+1-\alpha)R$$

Putting these together and cancelling out the common $R$ gives

$$u+1-\alpha={\alpha s_0\over u^2}+{(1-\alpha)s_1\over(u+1)^2}$$

Can you take it from there?