Proof: By Euclid

For let the cube number $A$ make the cube (number) $C$ (by) multiplying some number $B$.
I say that $B$ is cube.

fig05e

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