Proof: By Euclid
(related to Proposition: Prop. 8.12: Between two Cubes exist two Mean Proportionals)
 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 $D$.
 For let $C$ make $E$ (by) multiplying itself, and let it make $F$ (by) multiplying $D$.
 And let $D$ make $G$ (by) multiplying itself, and let $C$, $D$ make $H$, $K$, respectively, (by) multiplying $F$.
 And since $A$ is cube, and $C$ (is) its side, and $C$ has made $E$ (by) multiplying itself, $C$ has thus made $E$ (by) multiplying itself, and has made $A$ (by) multiplying $E$.
 And so, for the same (reasons), $D$ has made $G$ (by) multiplying itself, and has made $B$ (by) multiplying $G$.
 And since $C$ has made $E$, $F$ (by) multiplying $C$, $D$, respectively, thus as $C$ is to $D$, so $E$ (is) to $F$ [Prop. 7.17].
 And so, for the same (reasons), as $C$ (is) to $D$, so $F$ (is) to $G$ [Prop. 7.18].
 Again, since $C$ has made $A$, $H$ (by) multiplying $E$, $F$, respectively, thus as $E$ is to $F$, so $A$ (is) to $H$ [Prop. 7.17].
 And as $E$ (is) to $F$, so $C$ (is) to $D$.
 And thus as $C$ (is) to $D$, so $A$ (is) to $H$.
 Again, since $C$, $D$ have made $H$, $K$, respectively, (by) multiplying $F$, thus as $C$ is to $D$, so $H$ (is) to $K$ [Prop. 7.18].
 Again, since $D$ has made $K$, $B$ (by) multiplying $F$, $G$, respectively, thus as $F$ is to $G$, so $K$ (is) to $B$ [Prop. 7.17].
 And as $F$ (is) to $G$, so $C$ (is) to $D$.
 And thus as $C$ (is) to $D$, so $A$ (is) to $H$, and $H$ to $K$, and $K$ to $B$.
 Thus, $H$ and $K$ are two (numbers) in mean proportion to $A$ and $B$.
 So I say that $A$ also has to $B$ a cubed ratio with respect to (that) $C$ (has) to $D$.
 For since $A$, $H$, $K$, $B$ are four in (continued) proportion numbers, $A$ thus has to $B$ a cubed ratio with respect to (that) $A$ (has) to $H$ [Def. 5.10] .
 And as $A$ (is) to $H$, so $C$ (is) to $D$.
 And [thus] $A$ has to $B$ a cubed ratio with respect to (that) $C$ (has) to $D$.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"