Proof: By Euclid

For let the cube number $A$ measure the cube (number) $B$, and let $C$ be the side of $A$, and $D$ (the side) of $B$.
I say that $C$ measures $D$.

fig15e

For let $C$ make $E$ (by) multiplying itself.
And let $D$ make $G$ (by) multiplying itself.
And, further, [let] $C$ [make] $F$ (by) multiplying $D$, and let $C$, $D$ make $H$, $K$, respectively, (by) multiplying $F$.
So it is clear that $E$, $F$, $G$ and $A$, $H$, $K$, $B$ are in continued proportion in the ratio of $C$ to $D$ [Prop. 8.12].
And since $A$, $H$, $K$, $B$ are in continued proportion, and $A$ measures $B$, ($A$) thus also measures $H$ [Prop. 8.7].
And as $A$ is to $H$, so $C$ (is) to $D$.
Thus, $C$ also measures $D$ [Def. 7.20] .
And so let $C$ measure $D$.
I say that $A$ will also measure $B$.
For similarly, with the same construction, we can show that $A$, $H$, $K$, $B$ are in continued proportion in the ratio of $C$ to $D$.
And since $C$ measures $D$, and as $C$ is to $D$, so $A$ (is) to $H$, $A$ thus also measures $H$ [Def. 7.20] .
Hence, $A$ also measures $B$.
(Which is) the very thing it was required to show.
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