Proof: By Euclid

fig04e

For let $A$ make $D$ (by) multiplying itself.
Thus, $D$ is cube [Prop. 9.3].
And since $A$ has made $D$ (by) multiplying itself, and has made $C$ (by) multiplying $B$, thus as $A$ is to $B$, so $D$ (is) to $C$ [Prop. 7.17].
And since $A$ and $B$ are cube, $A$ and $B$ are similar solid (numbers) .
Thus, two numbers fall (between) $A$ and $B$ in mean proportion [Prop. 8.19].
Hence, two numbers will also fall (between) $D$ and $C$ in mean proportion [Prop. 8.8].
And $D$ is cube.
Thus, $C$ (is) also cube [Prop. 8.23].
(Which is) the very thing it was required to show.
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