\lim_{n\to \infty } \, n^3 \left(\sqrt[3]{n^6+n}-n^2\right)=\lim_{n\to \infty } \frac{\sqrt[3]{n^6\left(1+\frac{1}{n^5}\right)}-n^2}{\frac{1}{n^3}}=\lim_{n\to \infty }\frac{n^2\left(\sqrt[3]{\left(1+\frac{1}{n^5}\right)}-1\right)}{\frac{1}{n^3}}=\lim_{n\to \infty } \frac{\sqrt[3]{\left(1+\frac{1}{n^5}\right)}-1}{\frac{1}{n^5}}.