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Commentary on Proposition IX.6¶

If $a^2$ is a cube number, $a$ is also a cube.

For

$$1 : a = a : a^2 = a^2 : a^3.$$

Now $a^2$, $a^3$ are both cubes, and therefore “similar solids”; therefore there are two mean proportionals between them. [VIII. 19]

Therefore there are two mean proportionals between $a$, $a^2$. [VIII. 8]

And $a^2$ is a cube: therefore $a$ is also a cube number. [VIII. 23]

It will be noticed that the last step is not an exact quotation of the result of VIII. 23, because it is there the first of four terms which is known to be a cube, and the last which is proved to be a cube; here the case is reversed. But there is no difficulty. Without inverting the proportions, we have only to refer to VIII. 21 which proves that $a$, $a^2$, having two mean proportionals between them, are two similar solid numbers; whence, since $a^2$ is a cube, $a$ is also a cube.