$\underset{n}\lim{\Big(\sqrt[3\;]{\frac{\sin n}{n}+n^3-n^2}-\sqrt[3\;]{n^3+n}\Big)}$ $\underset{n}\lim{\Big(\sqrt[3\;]{\frac{\sin n}{n}+n^3-n^2}-\sqrt[3\;]{n^3+n}\Big)}$
I thought of using the squeeze theorem for$\;\frac{\sin n}{n},$ but I think it might incorrect here.
Another attempt was:
$$\underset{n}\lim{\frac{\frac{\sin n}{n}-n^2-n}{\sqrt[3\;]{\Big(\frac{\sin n}{n}+n^3-n^2}\Big)^2+\sqrt[3\;]{{\Big(\frac{\sin n}{n}+n^3-n^2}\Big)\Big(n^3+n}\Big)+\sqrt[3\;]{(n^3+n)^2}}}$$ I got stuck here.

Starting from Paramanand Singh's first equation
$$A=n\left(\sqrt[3]{1-\frac{1}{n}+\frac{\sin (n)} {n^4}}-\sqrt[3]{1+\frac{1}{n^2}}\right)$$ since $\sin(n)$ is bounded and very small when divided by $n^4$, consider, for the time being 
$$B=n\left(\sqrt[3]{1-\frac{1}{n}+\frac{a} {n^4}}-\sqrt[3]{1+\frac{1}{n^2}}\right)\quad \text{where} \quad -1 \leq a \leq 1$$ and use the binomial expansion or Taylor series (which I shall push to high order until $a$ appears.
This would give
$$B=-\frac{1}{3}-\frac{4}{9 n}-\frac{5}{81
   n^2}+\frac{81 a+17}{243 n^3}+O\left(\frac{1}{n^4}\right)$$
