Cube root of a line Well this may be simple but I am not getting it. 

Give a line segment (of length $l$)(and a segment of unit length if you require) how to construct a line of length $l^{1/3}$ with only a straight edge and compass?

I know how to draw a line with length $\sqrt l$ (through similarity process) but am at a loss at the cube root one. Can someone help?
 A: In terms of rational root theorem, it's not possible to draw cubed root of 2. However, I have found folks who figured things out using Euclidean geometry.
This one is the simplest:
https://demonstrations.wolfram.com/ConstructingTheCubeRootOfTwo/#more
Also, I came up with a hack myself. My approach is clunky and not precise past the thousandths decimal. I made this hack based upon some simple facts that

*

*cubed root of 2 is 1.25992105...

*sin(39.045) = 0.629930566

*2x(sin(39.045)) = 1.25986113

So if you had an isosceles triangle with the apex 78.09 degrees and the two sides are length 1, then the base should be 1.259.
Step 1: Create an angle that is .46875 degrees
Construct a right angle.
Trisect the right angle.
Take one of the 30 degree angles and bisect it 6 times. You will get an angle that is .46875 degrees.
Step 2: Create an angle that is 5.625 degrees.
Construct a right angle.
Bisect the right angle 4 times. You will get an angle that is 5.625 degrees
Step 3: Construct an angle that is 72 degrees.
Create a Golden Triangle (two sides equal to phi and the base is 1).
One of the base angles will be 72 degrees.
Step 4: Join angles from Step 1 - 3 together (Euclid Book 1 proposition 23). You will get an angle that is 78.09 degrees. Make a an isosceles triangle with sides = 1. The base will be very close to the cubed root of 2.
