Answer
$729q^6$
Work Step by Step
Using the $(r+1)$st term of the expansion of $(a+b)^n$, which is given by $\dfrac{n!}{(n-r)!r!}a^{n-r}b^r$, then the $
1
$st term of $
(3q-7r)^6
$ is
\begin{array}{l}
\dfrac{6!}{6!0!}(3q)^{6}(-7r)^{0}
\\\\=
1(729q^6)(1)
\\=
729q^6
\end{array}
