---
book: 9
number: 3
id: "IX.3"
kind: "theorem"
uses: ["[[book-8/proposition-8]]", "[[book-8/proposition-23]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:9.prop.3"
license: "CC-BY-SA-4.0"
---

# IX.3

*If a cube number by multiplying itself make some number, the product will be cube.*

## Proof

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

For let *C*, the side of *A*, be taken, and let *C* by multiplying itself make *D*.

It is then manifest that *C* by multiplying *D* has made *A*.

Now, since *C* by multiplying itself has made *D*, therefore *C* measures *D* according to the units in itself.

But further the unit also measures *C* according to the units in it; therefore, as the unit is to *C*, so is *C* to *D*. [[book-7/definitions#Definition 20|VII. Def. 20]]

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

But the unit also measures *C* according to the units in it; therefore, as the unit is to *C*, so is *D* to *A*.

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

Therefore between the unit and the number *A* two mean proportional numbers *C*, *D* have fallen in continued proportion.

Again, 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]]

But between the unit and *A* two mean proportional numbers have fallen; therefore two mean proportional numbers will also fall between *A*, *B*. [[book-8/proposition-8|VIII. 8]]

But, if two mean proportional numbers fall between two numbers, and the first be cube, the second will also be cube. [[book-8/proposition-23|VIII. 23]]

And *A* is cube; therefore *B* is also cube. Q. E. D.