A faster way to find the term $x^0$ in the expansion of $(1-\frac{x}{3})^{5}(1+\frac{2}{x})^{3}$ Now I am actually looking for a method in the general case when you have a product of two binomials. The method I use is equating coefficients; I have provided my example solution to this question as an answer. Now this is a pretty good method I reckon but I wonder if someone out there has a faster, more efficient solution to these types of questions that they would like to share (please?). The thing that annoys me is plugging huge sums of numbers into the calculator that takes more time than the answer itself.
I will not claim  this is faster ....
as $${(1+\frac{2}{x})}^3=\frac{{(x+2)}^3}{x^3}$$
we have to find the coefficient of $x^3$ in $$f(x)={(x+2)}^3{(1-x/3)}^5=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4..+a_8x^8$$
Notice that  $$f'''(x)=6a_3+24a_4x+..$$
$$f'''(0)=6a_3$$ From here can you finish it?
My way:

Find the general term for both binomials separately
To find the independent term, consider the possible expansions by inspection then plug in the appropriate values into the general terms
Use calculator to sum up values (very lengthy process if there are lots of cases)