Binomial expansion. Find the coefficient of $x$ in the expansion of $\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6$.
I've used the way that my teacher teach me. 
I've stuck in somewhere else. 
$\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6=\left(2-\frac{4}{x^3}\right)\left(x^6\left(1+\frac{2}{x^8}\right)^6\right)$
Can anyone teach me? Thanks in advance.

In case you're interested, here's another way of doing this kind of question: remove fractional terms to the outside first...
$$(2-\frac{4}{x^2})(x+\frac{2}{x^2})^6=\frac{1}{x^{14}}(2x^2-4)(x^3+2)^6$$
So now you just need the coefficient of $x^{15}$ in the expansion of $$(2x^2-4)(x^3+2)^6$$ towards which the $2x^2$ term makes no contribution, and we just get $$-4\times\binom65\times 2=-48$$
