How was this solution found? Consider an empty spherical bowl of radius $r$. I was trying to find the height to which I would need to fill the bowl with water so that it would be one quarter full (in terms of volume).
The total volume is $\frac{4}{3}\pi r^3$ and the volume filled be water up to a height $h$ is $\pi r h^2 - \frac{1}{3}\pi h^3$. For the bowl to be one quarter full (in terms of volume), $h$ needs to satisfy $h^3 - 3rh^2 + r^3 = 0$.
I couldn't solve this by hand and so used Maple, expecting only a numerical solution. Astonishingly, Maple was able to give two beautiful solutions. The positive one is such that
$$ \color{blue}{\frac{h}{r} = \sqrt{3}\sin\frac{\pi}{9}-\cos\frac{\pi}{9}+1}$$
I've tried making the substitution $h = \rho\mathrm{e}^{\mathrm{i}\theta}$ into $h^3 - 3rh^2 + r^3 = 0$ and then equating real and imaginary parts. Taking the imaginary parts gives $4\rho\cos^2\theta - 6r\cos\theta - \rho = 0$. This seemed promising, but solving doesn't give anything useful.
I can see that the solution could have been found by solving a general cubic and then simplifying the ghastly expressions by applying de Moivre's formula. 


*

*Is there an elegant and simple way of arriving at the result by hand?

*Under what conditions are such nice solutions possible?
 A: After some reading around, I think that I have found the solution. 
Starting with $h^3-3rh^2+r^3=0$, we make the substitution $h=x+r$ to get the reduced cubic $x^3 -3r^2x-r^3$, or equivalently $x^3 = 3r^2x+r^3$.
Next, we make the substitution $x=2r\cos\theta$ which yields $8r^3\cos^3\theta = 6r^3\cos\theta + r^3$. Since $r>0$ in the context of the original problem, we have $8\cos^3\theta = 6\cos\theta + 1$, or equivalently
$$4\cos^3\theta - 3\cos\theta = \frac{1}{2}$$
At this point we apply the identity $\cos3\theta \equiv 4\cos^3\theta  - 3\cos\theta$ to give $\displaystyle{\cos3\theta = \frac{1}{2}}$. This gives
$$3\theta = \pm\frac{\pi}{3},\pm\frac{5\pi}{3},\pm\frac{7\pi}{3},\pm\frac{11\pi}{3},\cdots$$
$$\theta = \pm\frac{\pi}{9},\pm\frac{5\pi}{9},\pm\frac{7\pi}{9},\pm\frac{11\pi}{9},\cdots$$
Since $x=2r\cos\theta$ and $h=x+r$, the solutions are
$$\color{blue}{h = 2r\cos\frac{\pi}{9}+r, \ 2r\cos\frac{5\pi}{9}+r, \ 2r\cos\frac{7\pi}{9}+r}$$
The negative solutions for $\theta$ are not used because $\cos\theta \equiv \cos(-\theta)$. The later solutions for $\theta$ are not used because, due to cosine's periodicity, they start to repeat the same answers.
Note that these three solutions take the same values as the ones in the OP.
A: Well I guess it depends if the solutions to depressed cubics can be counted as "getting there by hand" (by which I mean using the formula.. I'm not sure how hairy it is to be honest), but if they can, this is how I got to a depressed cubic:
$$h^3-3rh^2+r^3=0$$
$$h^2r(\frac{h}{r}-3)+r^3=0$$
$$\frac{h^2}{r^2}(\frac{h}{r}-3)+1=0$$
Set $x=\frac{h}{r}$
$$x^2(x-3)+1=0$$ 
$$x^3-3x^2+1=0$$
Since $(x-1)^3 = x^3-3x^2+3x-1$, we can write the above as 
$$(x-1)^3-3x+2=0$$
Set $z=x-1$
$$z^3-3z+5=0$$
Then I guess a formula would do from here.
