As the exponent 7 is odd, so there will be two middle terms in the expansion [3−

x3

6

]7
i.e., (

7+1

2

) th term and (

7+1

2

+1) th term i.e., 4th term & 5th term
∴ T4 = T3+1 =

7

‌

C3(3)7−3(−

x3

6

)3 = 35

(3)4

(6)3

= 35

(3)4

(6)3

(−x3)3 = 35 (

3

8

)(−x9) = -

105

8

x9 and T5 = T4+1 =

7

‌

C4(3)7−4(−

x3

6

)4 = 35

(3)3

(6)4

(−x3)4 =

35

48

x12