How can I prove that $8n^3 + 12n + 3\log^3n \neq \Omega (n^4)$ ?
I know that $8n^3 < 8n^4$ , $12n < 12n^4$ and $3\log^3n < 3n^4$ and then I can prove that $8n^3 + 12n + 3\log^3n = O(n^4)$ but it doesn't mean that it also $\neq \Omega (n^4)$ .
How should I prove it then?
Just do the division
$$\lim_{n\to\infty}{8n^3+12n+3\log^3n\over n^4}=0$$
since the limit exists, it is equal to the $\limsup$, which is exactly the proof it is not in $\Omega(n^4)$.
In your language, this means that if you assume there is $c>0$ such that $8n^3+12n+\log^3n\ge cn^4$ then this would imply
$$0< c=\lim_{n\to\infty}{cn^4\over n^4}\le \lim_{n\to\infty}{8n^3+12n+3\log^3n\over n^4}=0$$
a clear contradiction.