Square of Cube Number is Cube

Theorem

Let $a \in \N$ be a natural number.

Let $a$ be a cube number.

Then $a^2$ is also a cube number.

In the words of Euclid:

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

Proof 1

By the definition of cube number:

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

Thus:

 $\ds a^2$ $=$ $\ds \paren {k^3}^2$ $\ds$ $=$ $\ds k^6$ $\ds$ $=$ $\ds \paren {k^2}^3$

Thus:

$\exists r = k^2 \in \N: a = r^3$

Hence the result by definition of cube number.

$\blacksquare$

Proof 2

From Cube Number multiplied by Cube Number is Cube, if $a$ and $b$ are cube numbers then $a b$ is a cube number.

The result follows by setting $b = a$.

$\blacksquare$

Historical Note

This proof is Proposition $3$ 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.