Proof: By Euclid

fig03e

For let the side $C$ of $A$ have been taken.
And let $C$ make $D$ by multiplying itself.
So it is clear that $C$ has made $A$ (by) multiplying $D$.
And since $C$ has made $D$ (by) multiplying itself, $C$ thus measures $D$ according to the units in it [Def. 7.15] .
But, in fact, a unit also measures $C$ according to the units in it [Def. 7.20] .
Thus, as a unit is to $C$, so $C$ (is) to $D$.
Again, since $C$ has made $A$ (by) multiplying $D$, $D$ thus measures $A$ according to the units in $C$.
And a unit also measures $C$ according to the units in it.
Thus, as a unit is to $C$, so $D$ (is) to $A$.
But, as a unit (is) to $C$, so $C$ (is) to $D$.
And thus as a unit (is) to $C$, so $C$ (is) to $D$, and $D$ to $A$.
Thus, two numbers, $C$ and $D$, have fallen (between) a unit and the number $A$ in continued mean proportion.
Again, since $A$ has made $B$ (by) multiplying itself, $A$ thus measures $B$ according to the units in it.
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 two numbers have fallen (between) a unit and $A$ in mean proportion.
Thus two numbers will also fall (between) $A$ and $B$ in mean proportion [Prop. 8.8].
And if two (numbers) fall (between) two numbers in mean proportion, and the first (number) is cube, then the second will also be cube [Prop. 8.23].
And $A$ is cube.
Thus, $B$ is also cube.
(Which is) the very thing it was required to show.
∎

Thank you to the contributors under CC BY-SA 4.0!