# Why do third-octave bands have a ratio of cube root of two?

Without sounding tautological, I guess it boils down to the definition of the third octave, which is not: 2/3*Fc

but defined as the cube root of 2 ratio between frequencies.

Why is a third-octave not a "third of double the frequency" and instead "the cube root"? How does the square root play into this? Is there a log base 2 upon which this definition of a third octave is based?

To obtain the center frequency, it's the geometric mean of the upper and lower bounds, which gives you your addition sqrt 2, which gives you the upper bound Fc * 2^(1/6) = 2^(1/3) * sqrt(2) = 2^(1/3*1/2) and Fc / 2^(1/6) for the lower bound. Most tables start with 1kHz

A fixed interval corresponds to a fixed ratio of frequencies (at least in equal temperament, which is the most common tuning system in Western music). One octave is a factor of $2$. One semi-tone is a factor of $\sqrt[12]{2}$ because there are $12$ semi-tones in one octave. So one third of an octave (i.e., $4$ semi-tones) corresponds to a factor of $\sqrt[3]{2}$ because $\sqrt[3]{2}\cdot \sqrt[3]{2}\cdot \sqrt[3]{2}=2$ (i.e., one octave).