Answer
$-4060A^{3}$
Work Step by Step
$(a+b)^{n}=\left(\begin{array}{l}
n\\
0
\end{array}\right)a^{n}+\left(\begin{array}{l}
n\\
1
\end{array}\right)a^{n-1}b+\left(\begin{array}{l}
n\\
2
\end{array}\right)a^{n-2}b^{2}+\cdots+\left(\begin{array}{l}
n\\
n
\end{array}\right)b^{n}$
$\displaystyle \left(\begin{array}{l}
n\\
r
\end{array}\right)=\frac{n!}{r!(n-r)!},$
$n!=n(n-1)\cdot...\cdot 2\cdot 1,$
$n!=0$
Properties:
$\left(\begin{array}{l}
n\\
r
\end{array}\right)=\left(\begin{array}{l}
n\\
n-r
\end{array}\right), $
$ \left(\begin{array}{l}
k\\
r-1
\end{array}\right)+\left(\begin{array}{l}
k\\
r
\end{array}\right)=\left(\begin{array}{l}
k+1\\
r
\end{array}\right)$
--------
$...$In all the terms, exponents of a and b add to n...
In $(A+(-B))^{30} $, which has $31$ terms,
the last three terms contain powers of a:
28th, 29th, 30th, 31st:
$a^{3},\ \ \ a^{2},\ \ \ a^{1},\ \ \ a^{0}$
The 28th term in the expansion is where
the exponent of $A$ is $3,$
the exponent of $(-B)$ is $30-3=27,$ and
the coefficient is
$\left(\begin{array}{l}
30\\
27
\end{array}\right)= \left(\begin{array}{l}
30\\
30-27
\end{array}\right)= \displaystyle \left(\begin{array}{l}
30\\
3
\end{array}\right)=\frac{30\cdot 29\cdot 28}{1\cdot 2\cdot 3}=-4060$
The term is:
$-4060A^{3}(-B)^{27}=-4060A^{3}$
