Number divides Number iff Cube divides Cube

Theorem
Let $a, b \in \Z$.

Then:
 * $a^3 \mathrel \backslash b^3 \iff a \mathrel \backslash b$

where $\backslash$ denotes integer divisibility.

Proof
Let $a^3$ and $b^3$ be cube numbers.

From the corollary to Form of Geometric Progression of Integers:
 * $\left({a^3, a^2 b, a b^2, b^3}\right)$

is a geometric progression.

Let $a, b \in \Z$ such that $a^2 \mathrel \backslash b^2$.

Then from First Element of Geometric Progression that divides Last also divides Second:
 * $a^3 \mathrel \backslash a^2 b$

Thus:

Let $a \mathrel \backslash b$.

Then: