I figured this required the ratio test to do. First I have:
$$\lim_{n\to\infty} \left| \frac{2^{n+1}-3(n+1)^3}{1-4^{n+1}} \frac{1-4^n}{2^n -3n^3}\right |$$
Then I canceled out $2^n$ and $4^n$ from the top and bottom and was left with:
$$\lim_{n\to \infty} \left |\frac{2-3(n+1)^3}{4-3n^3}\right |$$
and got stuck.
Why not use limit comparison with the sequence $2^n/4^n=1/2^n$?
But if you're a glutton for punishment, we can use the Ratio Test:
$$ \left|\frac{2^{n+1}-3(n+1)^3}{1-4^{n+1}}\cdot\frac{1-4^n}{2^n-3n^3} \right|=\left|\frac{2^{n+1}-3(n+1)^3}{2^n-3n^3}\cdot\frac{4^n-1}{4^{n+1}-1} \right| $$ $$ =\left|\frac{2-3(n+1)^3 2^{-n}}{1-3n^3 2^{-n}}\cdot\frac{1-4^{-n}}{4-4^{-(n+1)}} \right| $$Worth showing: if $r>1$, $\lim_{n\to\infty} n^k/r^n=0$ for any $k$. Taking the limit, we have $$ \lim_{n\to\infty}\left|\frac{2-3(n+1)^3 2^{-n}}{1-3n^3 2^{-n}}\cdot\frac{1-4^{-n}}{4-4^{-(n+1)}} \right| = \frac{2-0}{1-0}\cdot \frac{1-0}{4-0}=\frac{1}{2}<1 $$