$\lim\sqrt[3]{n}=+\infty$
$\lim\dfrac{1}{\sqrt[3]{n}}=0$
$\lim\sqrt[3]{n^3-6n^2+1}=\lim n.\sqrt[3]{ 1-\dfrac{6}{n}+\dfrac{1}{n^3}}=+\infty$
$\lim(\sqrt[3]{n^3+1}-n)=\lim \dfrac{n^3+1-n^3}{\sqrt[3]{n^3+1}^2+n.\sqrt[3]{n^3+1}+n^2}=\lim\dfrac{1}{\sqrt[3]{n^3+1}^2+n.\sqrt[3]{n^3+1}+n^2}=\lim \dfrac{\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{1}{n^3}}^2+\sqrt[3]{1+\dfrac{1}{n^3}}+1}=0$