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.
 A: 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 $$-4\times\binom65\times 2=-48$$
A: Notice, we have 
$$\left(x+\frac{2}{x^2}\right)^6=^6C_0x^{6}\left(\frac{2}{x^2}\right)^{0}+^6C_1x^{5}\left(\frac{2}{x^2}\right)^{1}+^6C_2x^{4}\left(\frac{2}{x^2}\right)^{2}+^6C_3x^{3}\left(\frac{2}{x^2}\right)^{3}+^6C_4x^{2}\left(\frac{2}{x^2}\right)^{4}+^6C_5x^{1}\left(\frac{2}{x^2}\right)^{5}+^6C_6x^{0}\left(\frac{2}{x^2}\right)^{6}$$ Now, we have 
$$\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6$$
$$=2\left(x+\frac{2}{x^2}\right)^6-\frac{4}{x^3}\left(x+\frac{2}{x^2}\right)^6$$
Hence, the coefficient of $x$ in the above expansion $$=\text{coefficient of x in the expansion of}\ 2\left(x+\frac{2}{x^2}\right)^6-\text{coefficient of x in the expansion of}\ \frac{4}{x^3}\left(x+\frac{2}{x^2}\right)^6$$$$=2(0)-4(2\cdot ^6C_1)$$$$=-8(^6C_1)=-8(6)=-48$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{coefficient of}\ x=-48}}$$
