$ay^3 + xy = ab^3$, can I isolate $y$? I was wondering how much force it would take to compress a sphere of air (assuming Boyle's Law instead of the Real Gas laws, ignoring the engineering method of applying said force), so I started with atmospheric pressure * surface area + applied force = internal pressure * surface area. Using atmospheric pressure as 14.7 $lbs/in^2$, this becomes 14.7S + F = PS.
$S = 4πr^2$
P = P0 * V0 /V = 14.7*(r0/r$)^3$
$58.8πr^2 + F = 58.8πr$0$^3/r$
Defining the force applied as the independent variable "x," the final radius as the function "y," and setting the constants 58.8π = "a" and y0 = "b" for simplicity, gives
$ay^3 + xy = ab^3$
And now I'm stuck. I can calculate the initial radius of the sphere as a function of the force applied and of the final radius
b = ($y^3 + xy/a$)$^1$$^/$$^3$
and the force applied as a function of the two radii
x = a($b^3 - y^3$)/y
but I would also like to find the final radius "y" as a function of force "x" and initial radius "b." Is this possible?
 A: Instead of an exact answer (which might be hard to get some useful information from) we can instead obtain an approximation that tells us the behavior in different regimes.
Write the equation as
$$\left({y\over b}\right)^3 + \frac{x}{ab^2}\left({y\over b}\right) = 1$$
Now if the force is small in the sense that $\frac{x}{ab^2} \ll 1$ then trying a solution $y = b\left(1 - \epsilon\right)$ and expanding to first order in $\epsilon$ gives (where we have used $(1+\epsilon)^3 \approx 1+3\epsilon$)
$$y\approx b - \frac{x}{3ab}$$ 
Thus the radius decays lineary with the force.
Secondly, if the force is very large, ${x\over ab^2} \gg 1$ then trying a solution $\frac{y}{b} = \frac{ab^2}{x}\left(1 + \epsilon\right)$ and expanding to first order in $\epsilon$ gives us
$$\epsilon = -\frac{\left(\frac{ab^2}{x}\right)^3}{1 + 3\left(\frac{ab^2}{x}\right)^3} \approx - \left(\frac{ab^2}{x}\right)^3$$
so
$$y \approx \left(\frac{ab^2}{x} - \left(\frac{ab^2}{x}\right)^4\right)b \approx \frac{ab^3}{x}$$
Below I show the approximation togeather with the exact solution where I have choosen $y=b-\frac{x}{3ab}$ when $\frac{x}{ab^2} < 1.5$ and $y = \frac{ab^3}{x}$ when $\frac{x}{ab^2} > 1.5$ ($1.5$ is an arbritrary value chosen to get the best fit):

A: Beginning with $y^3+\frac{x}{a}y-b^3=0$,
 use Vieta substitution. 
$$y^3+\frac{x}{a}y-b^3$$Let $y=r-\frac{x}{3ar}$$$(r-\frac{x}{3ar})^3+\frac{x}{a}(r-\frac{x}{3ar})-b^3=0$$ $$r^3-\frac{xr}{a}+\frac{x^2}{3a^2r}-\frac{x^3}{27a^3r^3}+\frac{xr}{a}-\frac{x^2}{3a^2r}-b^3=0$$
$$r^6-b^3r^3-\frac{x^3}{27a^3}=0$$ Let $u=r^3$ $$u^2-b^3u-\frac{x^3}{27a^3}=0$$ Since your constants are nonnegative, this quadratic has a positive real root (by Descartes). Use this to recover a (physically useful) solution. $$y=\sqrt[3]{u}-\frac{x}{3a\sqrt[3]{u}}$$
