Cube Number multiplied by Cube Number is Cube

Theorem

Let $a, b \in \N$ be natural numbers.

Let $a$ and $b$ be cube numbers.

Then $a b$ is also a cube number.

In the words of Euclid:

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

(The Elements: Book $\text{IX}$: Proposition $4$)

Proof

By the definition of cube number:

$\exists r \in \N: r^3 = a$

$\exists s \in \N: s^3 = b$

Thus:

\(\ds a b\)

\(=\)

\(\ds \paren {r^3} \paren {s^3}\)

Power of Product

\(\ds \)

\(=\)

\(\ds \paren {r s}^3\)

Thus:

$\exists k = r s \in \N: a = k^3$

Hence the result by definition of cube number.

$\blacksquare$

Historical Note

This proof is Proposition $4$ of Book $\text{IX}$ of Euclid's The Elements.
The proof as given here is not that given by Euclid, as the latter is unwieldy and of limited use.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{IX}$. Propositions