---
book: 8
number: 12
id: "VIII.12"
kind: "theorem"
uses: ["[[book-7/proposition-17]]", "[[book-7/proposition-18]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:8.prop.12"
license: "CC-BY-SA-4.0"
---

# VIII.12

*Between two cube numbers there are two mean proportional numbers, and the cube has to the cube the ratio triplicate of that which the side has to the side.*

## Proof

Let *A*, *B* be cube numbers, and let *C* be the side of *A*, and *D* of *B*; I say that between *A*, *B* there are two mean proportional numbers, and *A* has to *B* the ratio triplicate of that which *C* has to *D*.

For let *C* by multiplying itself make *E*, and by multiplying *D* let it make *F*; let *D* by multiplying itself make *G*, and let the numbers *C*, *D* by multiplying *F* make *H*, *K* respectively.

Now, since *A* is a cube, and *C* its side, and *C* by multiplying itself has made *E*, therefore *C* by multiplying itself has made *E* and by multiplying *E* has made *A*.

For the same reason also *D* by multiplying itself has made *G* and by multiplying *G* has made *B*.

And, since *C* by multiplying the numbers *C*, *D* has made *E*, *F* respectively, therefore, as *C* is to *D*, so is *E* to *F*. [[book-7/proposition-17|VII. 17]]

For the same reason also, as *C* is to *D*, so is *F* to *G*. [[book-7/proposition-18|VII. 18]]

Again, since *C* by multiplying the numbers *E*, *F* has made *A*, *H* respectively, therefore, as *E* is to *F*, so is *A* to *H*. [[book-7/proposition-17|VII. 17]]

But, as *E* is to *F*, so is *C* to *D*.

Therefore also, as *C* is to *D*, so is *A* to *H*.

Again, since the numbers *C*, *D* by multiplying *F* have made *H*, *K* respectively, therefore, as *C* is to *D*, so is *H* to *K*. [[book-7/proposition-18|VII. 18]]

Again, since *D* by multiplying each of the numbers *F*, *G* has made *K*, *B* respectively, therefore, as *F* is to *G*, so is *K* to *B*. [[book-7/proposition-17|VII. 17]]

But, as *F* is to *G*, so is *C* to *D*; therefore also, as *C* is to *D*, so is *A* to *H*, *H* to *K*, and *K* to *B*.

Therefore *H*, *K* are two mean proportionals between *A*, *B*.

I say next that *A* also has to *B* the ratio triplicate of that which *C* has to *D*.

For, since *A*, *H*, *K*, *B* are four numbers in proportion, therefore *A* has to *B* the ratio triplicate of that which *A* has to *H*. [[book-5/definitions#Definition 10|V. Def. 10]]

But, as *A* is to *H*, so is *C* to *D*; therefore *A* also has to *B* the ratio triplicate of that which *C* has to *D*. Q. E. D.