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?
 A: The hint: 
Since $$a-b=\frac{a^3-b^3}{a^2+ab+b^2},$$ we obtain:
$$\left(\sqrt[3]{n^3+2n-1}-\sqrt{n^3+2n-3}\right)^6(1+3n+2n^3)^4=$$
$$=\left(\tfrac{n^3+2n-1-(n^3+2n-3)}{\sqrt[3]{(n^3+2n-1)^2}+\sqrt[3]{(n^3+2n-1)(n^3+2n-3)}+\sqrt[3]{n^3+2n-3)^2}}\right)^6(1+3n+2n^3)^4=$$
$$=\frac{64(1+3n+2n^3)^4}{\left(\sqrt[3]{(n^3+2n-1)^2}+\sqrt[3]{(n^3+2n-1)(n^3+2n-3)}+\sqrt[3]{n^3+2n-3)^2}\right)^6}=$$
$$=\frac{64\left(\frac{1}{n^3}+\frac{3}{n^2}+2\right)^4}{\left(\sqrt[3]{\left(1+\frac{2}{n^2}-\frac{1}{n^3}\right)^2}+\sqrt[3]{\left(1+\frac{2}{n^2}-\frac{1}{n^3}\right)\left(1+\frac{2}{n^2}-\frac{3}{n^3}\right)}+\sqrt[3]{\left(1+\frac{2}{n^2}-\frac{3}{n^3}\right)^2}\right)^6}.$$
A: 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.
A: Hint: Use that $$a^3-b^3=(a-b)(a^2+b^2+ab)$$
