---
book: 9
number: 6
id: "IX.6"
kind: "theorem"
uses: ["[[book-8/proposition-19]]", "[[book-8/proposition-8]]", "[[book-8/proposition-23]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:9.prop.6"
license: "CC-BY-SA-4.0"
---

# IX.6

*If a number by multiplying itself make a cube number, it will itself also be cube.*

## Proof

For let the number *A* by multiplying itself make the cube number *B*; I say that *A* is also cube.

For let *A* by multiplying *B* make *C*.

Since, then, *A* by multiplying itself has made *B*, and by multiplying *B* has made *C*, therefore *C* is cube.

And, since *A* by multiplying itself has made *B*, therefore *A* measures *B* according to the units in itself.

But the unit also measures *A* according to the units in it.

Therefore, as the unit is to *A*, so is *A* to *B*. [[book-7/definitions#Definition 20|VII. Def. 20]]

And, since *A* by multiplying *B* has made *C*, therefore *B* measures *C* according to the units in *A*.

But the unit also measures *A* according to the units in it.

Therefore, as the unit is to *A*, so is *B* to *C*. [[book-7/definitions#Definition 20|VII. Def. 20]]

But, as the unit is to *A*, so is *A* to *B*; therefore also, as *A* is to *B*, so is *B* to *C*.

And, since *B*, *C* are cube, they are similar solid numbers.

Therefore there are two mean proportional numbers between *B*, *C*. [[book-8/proposition-19|VIII. 19]]

And, as *B* is to *C*, so is *A* to *B*.

Therefore there are two mean proportional numbers between *A*, *B* also. [[book-8/proposition-8|VIII. 8]]

And *B* is cube; therefore *A* is also cube. cf. [[book-8/proposition-23|VIII. 23]] Q. E. D.