Golden mean in this equation Years ago I had started with this equation:
$2^{1/3}=(R/2)(\sqrt{1+8/R^3}-1)$
And arrived at the result
$2^{1/3}=\phi R$
Where phi, the golden ratio, is (sqrt(5) +1)/2. 
But at the moment can't retrace the steps that led to this result. I'm hoping someone can help me.
I took equation 5 from P. K. Aravind's pdf and adapted it to the more general case of a vertical orbital tether. I was looking at a vertical tether whose top throws payloads into parabolic trajectories.

 A: Start with
$$
2^{1/3}=\frac R2\left(\sqrt{1+\frac8{R^3}}-1\right)\tag1
$$
add $R$ to both sides
$$
2^{1/3}+R=\frac R2\left(\sqrt{1+\frac8{R^3}}+1\right)\tag2
$$
multiply $(1)$ and $(2)$:
$$
2^{2/3}+2^{1/3}R=\frac2R\tag3
$$
divide by $2^{1/3}R$
$$
\left(\frac{2^{1/3}}R\right)^2-\frac{2^{1/3}}R-1=0\tag4
$$
Thus, $\frac{2^{1/3}}R$ satisfies the same second order polynomial as $\phi$ and $-\frac1\phi$. Therefore, since $R\ge0$,
$$
R=\frac{2^{1/3}}\phi\tag5
$$
A: Let $x=\frac R 2$ and,for the time being consider, the equation
$$a=x\left(\sqrt{1+\frac 1 {x^3}}-1 \right)$$ Now, $x=\frac 1 y$ makes 
$$a y=\sqrt{1+ {y^3}}-1\implies (1+ay)^2=1+y^3$$ Expand it to get
$$a^2y^2+2ay=y^3\implies y(y^2-a^2y-2a)=0$$ So, if $y\neq 0$, this reduces to the quadratic the roots of which being 
$$y_\pm=\frac{1}{2} \left(a^2\pm\sqrt{a^3+8} \sqrt{a}\right)$$ Back to $x$ and replace $a$ by its value.
A: Replacing $R$ by its value,
$$2^{1/3}=(2^{1/3}/2\phi)(\sqrt{1+8\phi^3/2}-1),$$
$$2\phi=\sqrt{1+4\phi^3}-1,$$
$$(2\phi+1)^2=1+4\phi^3,$$
$$\phi^2+\phi=\phi^3,$$ which is right. You can backtrack.

If you want to solve from scratch, let $s:=1/R$ and
$$2^{4/3}s+1=\sqrt{1+8s^3},$$
$$8s^3-2^{8/3}s^2-2^{7/3}s=0.$$
Simplify by $4s$ and solve the quadratic equation.
