As the exponent 7 is odd, so there will be two middle terms in the expansion $\left[3-\frac{x^3}{6}\right]^7$ i.e., $\left(\frac{7+1}{2}\right)$ th term and $\left(\frac{7+1}{2}+1\right)$ th term i.e., 4th term & 5th term ∴ $T_4$ = $T_{3+1}$ = $\,{}^{7}C_{3}(3)^{7-3}\left(-\frac{x^3}{6}\right)^3$ = 35 $\frac{(3)^4}{(6)^3}$ = 35 $\frac{(3)^4}{(6)^3}(-x^3)^3$ = 35 $\left(\frac{3}{8}\right)(-x^9)$ = - $\frac{105}{8} x^9$ and $T_5$ = $T_{4+1}$ = $\,{}^{7}C_{4}(3)^{7-4}\left(-\frac{x^3}{6}\right)^4$ = 35 $\frac{(3)^3}{(6)^4}(-x^3)^4$ = $\frac{35}{48} x^{12}$