There exist two numbers in mean proportion to two (given)^{1} cube numbers. And (one) cube (number) has to the (other) cube (number) a cubed^{2} ratio with respect to (that) the side (of the former has) to the side (of the latter). * Let $A$ and $B$ be cube numbers, and let $C$ be the side of $A$, and $D$ (the side) of $B$. * I say that there exist two numbers in mean proportion to $A$ and $B$, and that $A$ has to $B$ a cubed ratio with respect to (that) $C$ (has) to $D$. * So I say that $A$ also has to $B$ a cubed ratio with respect to (that) $C$ (has) to $D$.
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Proofs: 1
Proofs: 1
In other words, between two given cube numbers there exist two numbers in continued proportion (translator's note). ↩
Literally, "triple" (translator's note). ↩