Most efficient way to approximately double a cube with ruler and compass? It is known that doubling a cube with ruler and compass is not possible. And this is because $\sqrt[3]{2}$ is not constructable. Mathologer visually demonstrates a proof of this in [1] for example. The obvious question then is, how close to this number can we get with ruler and compass? Is the answer arbitrarily close? How would we go about finding such a construction that gets to within $\epsilon$?

EDIT:
Since the answer to the first part of my question is "arbitrarily close", I'd like to ask what procedure gets within $\epsilon$ of $\sqrt[3]{2}$ in the least steps.
[1] https://www.youtube.com/watch?v=O1sPvUr0YC0
 A: Remark: This answer was written when the question only asked if there was some construction that gets arbitrarily close to $\sqrt[3]2$ (but posted after the question was edited). The question as it now appears, asking for efficient constructions using the fewest ruler and compass steps, is much much harder. I'm leaving the rest of this answer as is, for what it's worth.
Suppose we start with two points $A$ and $B$. We would like to get construct points arbitrarily close to the point $Z$ on the line through $A$ and $B$ such that $B$ is between $A$ and $Z$ and $AZ:AB=\sqrt[3]2:1\approx1.259921:1$. 
We start by drawing the line through $A$ and $B$ and then a circle centered at $B$ of radius $AB$, which produces a point $C$. Note that $B$ lies between $A$ and $C$ and $AC:AB=2:1\gt\sqrt[3]2:1$.
We can now bisect the line segment $\overline{BC}$ by drawing a circle centered at $C$ of radius $BC=AB$. This circle intersects the previous circle (from the previous paragraph) at two points $P$ and $Q$. Drawing the line through $P$ and $Q$, we obtain the midpoint of the line segment $\overline{BC}$. Call this point $D$.
We find that $AD:AB=1.5:1\gt\sqrt[3]2:1$, so we next bisect $BD$, obtaining the midpoint $E$, for which $AE:AB=1.25:1\lt\sqrt[3]2:1$.  The comparison tells us to bisect $ED$ next, obtaining a midpoint $F$ for which we find $AF:AB=1.375:1\gt\sqrt[3]2:1$. Successive bisections will, over time, eventually get us as close to the desired (but nonconstructible) point $Z$ as we like. The total number of steps in the construction is $O(|\log\epsilon|)$; roughly speaking, each bisection gives the next bit in the binary expansion of $\sqrt[3]2$, and each bisection takes either two or three new ruler-and-compass steps.
It might be of interest to see if there are quicker ways to construct two points, say $X$ and $Y$, such that $XY:AB\approx\sqrt[3]2:1$, where "quicker" is determined by the total number of circles and lines that need to be drawn, starting from the given points $A$ and $B$. There are certainly quicker numerical methods for approximating a cube root (e.g., Newton-Raphson), but it's not clear (to me, at least) how easily they translate into an efficient sequence of ruler and compass steps. Put differently, if you restrict yourself to a total of $N$ ruler and compass steps, what's the closest you can get to the cube root of $2$? It's conceivable that the best construction with $N+1$ steps is entirely different than the best construction with $N$ steps.
