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SVppose that A being a cube number multiplieng himselfe, do produce the num∣ber B. Then I say that B is a cube number. Take the side of A, and let the same be the number C, and let C multiplieng himselfe produce the number D. Now it is manifest that C multiplieng D produceth A (by the 20. definition of the seuēth)
which are in C. But vnitie also measureth C by those vnities which are in C. Wherfore as vnitie is to C, so is C to D. Againe forasmuch as C multiplieng D produceth A: therefore D measureth A by those vnities which are in C. But vnitie measureth C by those vnities which are in C: wherefore as vnitie is to C, so is D to A. But as vnitie is to C, so is C to D, wherfore as vnitie is to C, so is C to D & D to A. Wherefore betwene vnitie & A there are two meane proportionall numbers, namely, C, D. Againe forasmuch as A multiplieng him∣selfe produced B, therefore A measureth B by those vnities which are in A. But vnitie also measureth A by those vnities which are in A. Wherfore as vnitie is to A, so is A to B. But be∣twene A and vnitie, there are two meane proportionall numbers. Wherfore betwene A and B also there are two meane proportionall numbers by the 8. of the eight. But if betwene two numbers, there be two meane proportionall numbers, and if the first be a cube number, the fourth also shall be a cube number by the 21. of the eight. But A is a cube number, wherefore B also is a cube number which was required to be proued.
Demonstra∣tion.