If two numbers have to one another the ratio which a cubic number has to a cubic number, and the first is a cube, then the second is also a cube.

Let the two numbers *A* and *B* have to one another the ratio which the cubic number *C* has to the cubic number *D,* and let *A* be a cube.

I say that *B* is also a cube.

Since *C* and *D* are cubes, *C* and *D* are similar solid numbers, therefore two mean proportional numbers fall between *C* and *D.*

Since as many numbers fall in continued proportion between those which have the same ratio with *C* and *D* as fall between *C* and *D,* therefore two mean proportional numbers *E* and *F* fall between *A* and *B.*

Since, then, the four numbers *A, E, F,* and *B* are in continued proportion, and *A* is a cube, therefore *B* is also a cube.

Therefore, *if two numbers have to one another the ratio which a cubic number has to a cubic number, and the first is a cube, then the second is also a cube.*

Q.E.D.

This proposition is used in IX.10.