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