Two numbers fall (between) two similar solid numbers in mean proportion. And a solid (number) has to a similar solid (number) a cubed^{1} ratio with respect to (that) a corresponding side (has) to a corresponding side. * Let $A$ and $B$ be two similar solid numbers, and let $C$, $D$, $E$ be the sides of $A$, and $F$, $G$, $H$ (the sides) of $B$. * And since similar solid (numbers) are those having proportional sides [Def. 7.21] , thus as $C$ is to $D$, so $F$ (is) to $G$, and as $D$ (is) to $E$, so $G$ (is) to $H$. * I say that two numbers fall (between) $A$ and $B$ in mean proportion, and (that) $A$ has to $B$ a cubed ratio with respect to (that) $C$ (has) to $F$, and $D$ to $G$, and, further, $E$ to $H$.
(not yet contributed)
Proofs: 1
Literally "triple" (translator's note). ↩