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

It is known that in the expansion of $$(a+b)^n$$ if n is odd, then there are two middle terms namely $$\left (\dfrac{n+1}{2} \right )^{th}$$ term and $$\left (\dfrac{n+1}{2} +1 \right )^{th}$$ term.
Therefore, the middle terms in the expansion of $$\left (3- \dfrac{x^3}{6}  \right )^7$$ are 
$$\left (\dfrac{7+1}{2}\right)=4^{th}$$ term and $$\left (\dfrac{7+1}{2} +1 \right)=5^{th}$$ term.
$$T_4 = T_{3+1}= ^7C_3(3)^{7-3}\left ( -\dfrac{x^3}{6} \right )^3= (-1)^3 \dfrac{7!}{3!4!}.3^4.\dfrac{x^9}{6^3}$$
$$= - \dfrac{7.6.5.4!}{3.2.4!}.3^4.\dfrac{1}{2^3.3^3}.x^9=-\dfrac{105}{8}x^9$$
$$ T_5 = T_{4+1} = ^7C_4(3)^{7-4}\left ( -\dfrac{x^3}{6} \right )^4= (-1)^4 \dfrac{7!}{4!3!}.(3)^3.\dfrac{x^{12}}{6^4}$$
$$= \dfrac{7.6.5.4!}{4!3.2}\dfrac{3^3}{2^4.3^4}.x^{12}=\dfrac{35}{48}x^{12}$$
Thus, the middle terms in the expansion of $$\left (3- \dfrac{x^3}{6}  \right )^7$$ are 
$$-\dfrac {105}{8}x^9$$ and $$\dfrac {35}{48}x^{12}$$