College Algebra, Chapter 9, Review Exercises, Section Review Exercises, Problem 80

Determine the first three terms in the expression of $\displaystyle \left( b^{\frac{-2}{3}} + b^{\frac{1}{3}} \right)^{20}$
The $n$th term of the binomial expansion $(a+b)^n$ is defined as
$\displaystyle
\left(
\begin{array}{c}
n \\
r -1
\end{array}
\right)
(a)^{n-r+1} (b)^{r-1}
\qquad
\text{or}
\qquad
nC_{r-1} = (a)^{n-r+1} (b)^{r-1}$

If we rewrite the expression as $\displaystyle \left( b^{\frac{1}{3}} + b^{\frac{-2}{3}} \right)^{20}$ we have, $n = 20$, $a = b^{\frac{1}{3}}$ and $b = b^{\frac{-2}{3}}$. So the first term is


$
\begin{equation}
\begin{aligned}

&=
\left(
\begin{array}{c}
20 \\
1-1
\end{array}
\right)

\left(
b^{\frac{1}{3}}
\right)^{20 - 1 +1}

\left(
b^{\frac{-2}{3}}
\right)^{1-1}
\\
\\

&=
\left(
\begin{array}{c}
20 \\
0
\end{array}
\right)

\left(
b^{\frac{1}{3}}
\right)^{20}

\left(
b^{\frac{-2}{3}}
\right)^{0}
\\
\\
&=
b^{\frac{20}{3}}

\end{aligned}
\end{equation}
$


The second term is

$
\begin{equation}
\begin{aligned}

&=
\left(
\begin{array}{c}
20 \\
2-1
\end{array}
\right)

\left(
b^{\frac{1}{3}}
\right)^{20 - 2 +1}

\left(
b^{\frac{-2}{3}}
\right)^{2-1}
\\
\\

&=
\left(
\begin{array}{c}
20 \\
1
\end{array}
\right)

\left(
b^{\frac{1}{3}}
\right)^{19}

\left(
b^{\frac{-2}{3}}
\right)
\\
\\
&=
\left(
\begin{array}{c}
20 \\
1
\end{array}
\right)

\left(
b^{\frac{19}{3}}
\right)

\left(
b^{\frac{-2}{3}}
\right)
\\
\\
&=
\left(
\begin{array}{c}
20 \\
1
\end{array}
\right)

(b)^{\frac{19}{3} - \frac{2}{3}}
\\
\\
&= 20b^{\frac{17}{3}}
\end{aligned}
\end{equation}
$


Then, the third term is

$
\begin{equation}
\begin{aligned}

&=
\left(
\begin{array}{c}
20 \\
3-1
\end{array}
\right)

\left(
b^{\frac{1}{3}}
\right)^{20 - 3 +1}

\left(
b^{\frac{-2}{3}}
\right)^{3-1}
\\
\\

&=
\left(
\begin{array}{c}
20 \\
2
\end{array}
\right)

\left(
b^{\frac{1}{3}}
\right)^{18}

\left(
b^{\frac{-2}{3}}
\right)^{2}
\\
\\

&=
\left(
\begin{array}{c}
20 \\
2
\end{array}
\right)

\left(
b^{\frac{18}{3}}
\right)
\left(
b^{\frac{-4}{3}}
\right)
\\
\\

&=
\left(
\begin{array}{c}
20 \\
2
\end{array}
\right)

\left(
b^{\frac{18}{3}- \frac{4}{3}}
\right)
\\
\\
&= 190b^{\frac{14}{3}}

\end{aligned}
\end{equation}
$

Therefore, the first three terms are,
$b^{\frac{20}{3}} + 20b^{\frac{17}{3}}$ and $190b^{\frac{14}{3}}$ respectively.