Proof: By Euclid

Let $A$ and $B$ be similar solid numbers.
I say that $A$ has to $B$ the ratio which (some) cube number (has) to a(nother) cube number.

fig27e

For since $A$ and $B$ are similar solid (numbers) , two numbers thus fall (between) $A$ and $B$ in mean proportion [Prop. 8.19].
Let $C$ and $D$ have (so) fallen.
And let the least numbers, $E$, $F$, $G$, $H$, having the same ratio as $A$, $C$, $D$, $B$, (and) equal in multitude to them, have been taken [Prop. 8.2].
Thus, the outermost of them, $E$ and $H$, are cube [Prop. 8.2 corr.] .
And as $E$ is to $H$, so $A$ (is) to $B$.
And thus $A$ has to $B$ the ratio which (some) cube number (has) to a(nother) cube number.
(Which is) the very thing it was required to show.
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