Proof: By Euclid

For let two numbers, $A$ and $B$, have to one another the ratio which the cube number $C$ (has) to the cube number $D$.
And let $A$ be cube.
So I say that $B$ is also cube.

fig25e

For since $C$ and $D$ are cube (numbers), $C$ and $D$ are (thus) similar solid (numbers) .
Thus, two numbers fall (between) $C$ and $D$ in mean proportion [Prop. 8.19].
And as many (numbers) as fall in between $C$ and $D$ in continued proportion, so many also (fall) in (between) those (numbers) having the same ratio as them (in continued proportion) [Prop. 8.8].
And hence two numbers fall (between) $A$ and $B$ in mean proportion.
Let $E$ and $F$ (so) fall.
Therefore, since the four numbers $A$, $E$, $F$, $B$ are in continued proportion, and $A$ is cube, $B$ (is) thus also cube [Prop. 8.23].
(Which is) the very thing it was required to show.
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