Solution to a cubic-like polynomial with fractional powers I wish to solve the following equation for x, but am not certain how:
$\frac{(1+x)^{1/3}}{1+x^{1/3}} = (\frac{3^{1/3}}{.691})(R/a)(p/h)^{-1/3}$
I suspect that there is some way to manipulate this into a cubic equation and solve by the usual methods, but despite much work I have not succeeded in finding any solution for x. Any help would be appreciated.

Set $x^{1/3}=z\to x=z^3$
and $k=(\frac{3^{1/3}}{.691})(R/a)(p/h)^{-1/3}$
the equation becomes
$$\frac{\sqrt[3]{z^3+1}}{z+1}=k$$
$$\frac{z^3+1}{(z+1)^3}=k^3$$
$$\frac{(z+1)(z^2-z+1)}{(z+1)^3}=k^3$$
$$z^2-z+1=k^3(z^2+2z+1)$$
$$z=\frac{-2 k^3-1\pm\sqrt{12 k^3-3}}{2 \left(k^3-1\right)}$$
$$x=z^3\to x=\left(\frac{-2 k^3-1\pm\sqrt{12 k^3-3}}{2 \left(k^3-1\right)}\right)^3$$
set back the value of $k$
$$x_1=\frac{0.125 \left(a^3 p \left(\sqrt{\frac{36.3703 h R^3}{a^3 p}-3}+1\right)+6.06172 h R^3\right)^3}{\left(a^3 p-3.03086 h R^3\right)^3}$$
$$x_2=-\frac{0.125 \left(a^3 p \left(\sqrt{\frac{36.3703 h R^3}{a^3 p}-3}-1\right)-6.06172 h R^3\right)^3}{\left(a^3 p-3.03086 h R^3\right)^3}$$
