Calculate limit with squares $\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4$ 
$$\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4 $$

What I did was to multiply it and I got $\frac{1}{2}$ as the final result. Could someone confirm if it's correct?

Use the binomial expansion or, better, Taylor series to show that
$$\sqrt[3]{n^3+2n-1}=n+\frac{2}{3 n}-\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\sqrt[3]{n^3+2n-3}=n+\frac{2}{3 n}-\frac{1}{ n^2}+O\left(\frac{1}{n^3}\right)$$
$$\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}=\frac{2}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ The remaining looks to be simple.
