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.
 A: 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?
A: Using the binomial theorem,
\begin{align*}
\left( 1 - \frac{x}{3} \right)^5 &= \sum_{i=0}^{5} \binom{5}{i} 1^{5-i} \left(\frac{-x}{3} \right)^i  \\
    &= \sum_{i=0}^5 \binom{5}{i} (-3)^{-i} x^i  \text{ and }  \\
\left( 1 + \frac{2}{x} \right)^3 &= \sum_{j=0}^{3} \binom{3}{j}1^{3-j} \left(\frac{2}{x} \right)^j  \\
    &= \sum_{j=0}^{3} \binom{3}{j} 2^j x^{-j}
\end{align*}
So \begin{align*}
\left( 1 - \frac{x}{3} \right)^5 \left( 1 + \frac{2}{x} \right)^3
    &= \left( \sum_{i=0}^5 \binom{5}{i} (-3)^{-i} x^i \right) \left( \sum_{j=0}^{3} \binom{3}{j} 2^j x^{-j} \right)  \\
    &= \sum_{i=0}^5 \sum_{j=0}^{3} \binom{5}{i} \binom{3}{j} (-3)^{-i} 2^j x^{i-j}
\end{align*}
The only contributions to $x^0$ have $i - j = 0$, so $i = j$.  The range of $i$ is $[0,5]$ and the range of $j$ is $[0,3]$.  The intersection of these two intervals is $[0,3]$, so we have four terms contributing to the coefficient of $x^0$:
$$  \sum_{j=0}^{3} \binom{5}{j} \binom{3}{j} (-3)^{-j} 2^j  $$
(It seems that we have done a lot of work to get here.  With practice, you can jump straight to this expression.)
Call the four terms $a_j = \binom{5}{j} \binom{3}{j} (-3)^{-j} 2^j$ for $j \in [0,3]$.  Notice
$$ a_0 = \binom{5}{0} \binom{3}{0} (-3)^{-0} 2^0 = 1  \text{.}  $$
Then
\begin{align*}
a_1 &= a_0 \cdot \frac{5}{1} \cdot \frac{3}{1} \cdot \frac{-1}{3} \cdot 2 = -10  \text{,}  \\
a_2 &= a_1 \cdot \frac{4}{2} \cdot \frac{2}{2} \cdot \frac{-1}{3} \cdot 2 = \frac{40}{3}  \text{, and}  \\
a_3 &= a_2 \cdot \frac{3}{3} \cdot \frac{1}{3} \cdot \frac{-1}{3} \cdot 2 = \frac{-80}{27}  
\end{align*}
This says we can compute the next term from the previous one.  The second and third factors are the old mental arithmetic method for calculating rows of Pascal's triangle by multiplying by successive fractions having decreasing numerators and increasing denominators.
Then the coefficient of $x^0$ is
$$  1 - 10 + \frac{40}{3} - \frac{80}{27} = \frac{37}{27}  \text{.}  $$
A: It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.

We obtain
\begin{align*}
\color{blue}{[x^0]}&\color{blue}{\left(1+\frac{2}{x}\right)^3\left(1-\frac{x}{3}\right)^5}\\
&=[x^0]\sum_{j=0}^3\binom{3}{j}2^jx^{-j}\left(1-\frac{x}{3}\right)^5\tag{1}\\
&=\sum_{j=0}^3\binom{3}{j}2^j[x^j]\left(1-\frac{x}{3}\right)^5\tag{2}\\
&\,\,\color{blue}{=\sum_{j=0}^3\binom{3}{j}\binom{5}{j}2^j\left(-\frac{1}{3}\right)^j}\tag{3}
\end{align*}
and the result can be manually calculated by summing up $4$ terms.

Comment:

*

*In (1) we apply the binomial theorem to the left-hand binomial.


*In (2) we use the linearity of the coefficient of operator and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.


*In (3) we select the coefficient of $x^j$.
A: 
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)

