Prove the cuberoot of 2 is irrational

This is not, probably, the most convincing or explanatory proof, and this certainly does not answer the question, but I love this proof.

Suppose that $ \sqrt[3]{2} = \frac p q $. Then $ 2 q^3 = p^3 $. This means $ q^3 + q^3 = p^3 $. The last equation has no nontrivial integer solutions due to Fermat's Last Theorem.

If $p$ is prime, and $p\mid a_1a_2\cdots a_n$ then $p\mid a_i$ for some $i$.

Now, let $p=2$, $n=3$ and $a_i=a$ for all $i$.

Your proof is fine, once you understand that step 6 implies step 7:

This is simply the fact odd $\times$ odd $=$ odd. (If $a$ were odd, then $a^3$ would be odd.)

Anyway, you don't need to assume that $a$ and $b$ are coprime:

Consider $2b^3 = a^3$. Now count the number of factors of $2$ on each side: on the left, you get an number of the form $3n+1$, while on the right you get an a number of the form $3m$. These numbers cannot be equal because $3$ does not divide $1$.