On the convergence of the series $ \sum_n \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} $ I've got some problem with this series
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)}
$$
I know that the series diverges by a comparison test, but I've got a lot of trouble to prove it. I've tried some algebraic tricks like 
$$
a - b = \frac{a^{3} - b^{3}}{a^{2} + ab + b^{2}},
$$
where $ a = \sqrt[3]{(n^{3}+n)} $ and $ b = \sqrt[3]{(n^3-n)} $.
It gave me
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} = \sum_{n=1}^{\infty} \frac{2n}{\sqrt[3]{(n^{3}+n)^{2}} + \sqrt[3]{n^{6} - n^{2}} + \sqrt[3]{(n^3-n)^{2}}}
$$
but for me it looks like a blind valley. I would appreciate every help, thank you.

Your idea was good:
$$\sqrt[3]{(n^3 + n)^2} = n^2\sqrt[3]{(1 + 1/n^2)^2},$$
$$\sqrt[3]{n^6 - n^2} = n^2\sqrt[3]{1 - 1/n^4},$$
$$\sqrt[3]{(n^3 - n)^2} = n^2\sqrt[3]{(1 - 1/n^2)^2}.$$
So
$$
\sqrt[3]{(n^3 + n)^2} + \sqrt[3]{n^6 - n^2} + \sqrt[3]{(n^3 - n)^2}\sim 3n^2
$$
and
$$
\frac{2n}{\sqrt[3]{(n^3 + n)^2} + \sqrt[3]{n^6 - n^2} + \sqrt[3]{(n^3 - n)^2}}\sim\frac{2}{3n}.
$$
