How to solve $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$ without L'Hopital? $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$
$\lim_{n \to \infty}\frac{1}{\sqrt[6]{(n^3+n+1)^2}-\sqrt[6]{(n^2-n+2)^3}}$ but because this limit is still the type of $\frac{1}{\infty-\infty}$ I tried to do this:
$\lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{(n^3+n+1)^2-(n^2-n+2)^3} = \lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{3n^5-7n^4+15n^3-17n^2+14n-7}$ I'm totally stuck here. I would divide the fraction by $3n^5$ and then the solution is $0$. Not the correct answer.
Did I miss something?

fast solution,as another idea$$\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}=\\
\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+0n^2+n+1}-\sqrt{n^2-n+2}}=\\
\lim_{n \to \infty}\frac{1}{\sqrt[3]{(n+0)^3-0+0n^2+n+1}-\sqrt{(n-\frac12)^2-\frac14+2}}=\\
\lim_{n \to \infty}\frac{1}{n-(n-\frac12)}=\\2$$
REMARK:
$$n^3+an^2+bn+c=\\(n+\frac a3)^3-(3n^2.\frac a3+3n(\frac  a3)^2+(\frac a3)^3)+bn+c\\(n\to \infty) \implies n^3+an^2+bn+c\sim (n+\frac a3)^3 $$ so $$n^3+n+1=(n+0)^3-0^3+n+1$$ also for $n^2+an+b=(n+\frac a2)^2-(\frac a2)^2+c$
