Proof: By Euclid
(related to Proposition: Prop. 9.03: Square of Cube Number is Cube)
 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 BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"