Proof: By Euclid

fig06e

For let $A$ make $C$ (by) multiplying $B$.
Therefore, since $A$ has made $B$ (by) multiplying itself, and has made $C$ (by) multiplying $B$, $C$ is thus cube.
And since $A$ has made $B$ (by) multiplying itself, $A$ thus measures $B$ according to the units in ($A$).
And a unit also measures $A$ according to the units in it.
Thus, as a unit is to $A$, so $A$ (is) to $B$.
And since $A$ has made $C$ (by) multiplying $B$, $B$ thus measures $C$ according to the units in $A$.
And a unit also measures $A$ according to the units in it.
Thus, as a unit is to $A$, so $B$ (is) to $C$.
But, as a unit (is) to $A$, so $A$ (is) to $B$.
And thus as $A$ (is) to $B$, (so) $B$ (is) to $C$.
And since $B$ and $C$ are cube, they are similar solid (numbers) .
Thus, there exist two numbers in mean proportion (between) $B$ and $C$ [Prop. 8.19].
And as $B$ is to $C$, (so) $A$ (is) to $B$.
Thus, there also exist two numbers in mean proportion (between) $A$ and $B$ [Prop. 8.8].
And $B$ is cube.
Thus, $A$ is also cube [Prop. 8.23].
(Which is) the very thing it was required to show.
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