$\underset{n}\lim{\Big(\sqrt[3\;]{\frac{\sin n}{n}+n^3-n^2}-\sqrt[3\;]{n^3+n}\Big)}$ $\underset{n}\lim{\Big(\sqrt[3\;]{\frac{\sin n}{n}+n^3-n^2}-\sqrt[3\;]{n^3+n}\Big)}$
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.
 A: You can definitely use a squeeze theorem argument here, but other than getting rid of the sine, it doesn't help much, and then you'd need to evaluate two limits instead of just one!
Anyway, your second attempt looks like a good start.  
Next step is to factor out $n^2$ from the denominator and then divide the numerator by said $n^2$.  It should then have a form you can evaluate directly (here, using a simple squeeze theorem on each term with a sine).
A: You can take $n^3$ (term of highest degree) out from the radical symbol to get the expression $$n\left(\sqrt[3]{1-\frac{1}{n}+\frac{\sin n} {n^4}}-\sqrt[3]{1+\frac{1}{n^2}}\right)$$ This is clearly of the form $$n(A^{1/3}-B^{1/3})$$ where $A, B$ tend to $1$. And this can be further rewritten as $$n(A-1)\cdot\frac{A^{1/3}-1}{A-1}-n(B-1)\cdot\frac{B^{1/3}-1}{B-1}$$ Note that $n(A-1)\to - 1, n(B-1)\to 0$ and the two fractions above each tend to $1/3$, therefore the desired limit is $-1/3$.

In general try to avoid rationalization for roots more complicated than square roots. The key is to never type/write complicated expressions because they are a great distraction to the eye and mind. Always go for symbols (like $A, B$ used here) and then you can think clearly. 
A: 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
$$B=-\frac{1}{3}-\frac{4}{9 n}-\frac{5}{81
   n^2}+\frac{81 a+17}{243 n^3}+O\left(\frac{1}{n^4}\right)$$
