Number does not divide Number iff Cube does not divide Cube

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

Then:
 * $a \nmid b \iff a^3 \nmid b^3$

where $a \nmid b$ denotes that $a$ is not a divisor of $b$.

Proof
Let $a \nmid b$.


 * $a^3 \mathrel \backslash b^3$
 * $a^3 \mathrel \backslash b^3$

where $\backslash$ denotes divisibility.

Then by Number divides Number iff Cube divides Cube:
 * $a \mathrel \backslash b$

From Proof by Contradiction it follows that $a^2 \mathrel \backslash b^2$ is false.

Thus $a^3 \nmid b^3$.

Let $a^3 \nmid b^3$.


 * $a \mathrel \backslash b$

Then by Number divides Number iff Cube divides Cube:
 * $a^3 \mathrel \backslash b^3$

From Proof by Contradiction it follows that $a \mathrel \backslash b$ is false.

Thus $a \nmid b$.