Find the middle terms in the expansions of $\left(3-\frac{x^{3}}{6}\right)^{7}$

It is known that in the expansion of $(a+b)^{n},$ in $n$ is odd, then there are two middle terms, Namely $\left(\frac{n+1}{2}\right)^{th}$ term and $\left(\frac{n+1}{2}+1\right)^{th}$ term

Therefore, the middle terms in the expansion $\left(3-\frac{x^{3}}{6}\right)^{7}$ are $\left(\frac{7+1}{2}\right)^{th}=4^{th}$ and $\left(\frac{7+1}{2}+1\right)^{th}=5^{th}$ term

${T_4} = {T_{3 + 1}} = {\,^7}{C_3}{(3)^{7 - 3}}{\left( { - \frac{{{x^3}}}{6}} \right)^3} = {( - 1)^3}\frac{{7!}}{{3!4!}} \cdot {3^4} \cdot \frac{{{x^9}}}{{{6^3}}}$

$=-\frac{7 \cdot 6 \cdot 5 \cdot 4 !}{3 \cdot 2 \cdot 4 !} \cdot 3^{4} \cdot \frac{1}{2^{3} \cdot 3^{3}} \cdot x^{9}=-\frac{105}{8} x^{9}$

${T_5} = {T_{4 + 1}} = {\,^7}{C_4}{(3)^{7 - 4}}{\left( { - \frac{{{x^3}}}{6}} \right)^4} = {( - 1)^4}\frac{{7!}}{{4!3!}} \cdot {3^3} \cdot \frac{{{x^{12}}}}{{{6^4}}}$

$=\frac{7 \cdot 6 \cdot 5.4 !}{4 ! \cdot 3 \cdot 2} \cdot \frac{3^{3}}{2^{4} \cdot 3^{4}} \cdot x^{12}=\frac{35}{48} x^{12}$

Thus, the middle terms in the expansion of $\left(3-\frac{x^{3}}{6}\right)^{7}$ are $-\frac{105}{8} x^{9}$ and $\frac{35}{48} x^{12}$