(related to Proposition: Prop. 9.05: Number multiplied by Cube Number making Cube is itself Cube)

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

- 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.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"