Binomial Theorem: Term Independent of x

The general term in $(a + b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Here $n = 9, a = \frac{3}{2}x^2, b = -\frac{1}{3x}$: $T_{r+1} = \binom{9}{r} \left( \frac{3}{2}x^2 \right)^{9-r} \left( -\frac{1}{3x} \right)^r$ $= \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} (-1)^r \left(\frac{1}{3}\right)^r (x^2)^{9-r} (x^{-1})^r$ $= \binom{9}{r} (-1)^r \frac{3^{9-r}}{2^{9-r} 3^r} x^{18 - 2r - r} = \binom{9}{r} (-1)^r \frac{3^{9-2r}}{2^{9-r}} x^{18 - 3r}$

For the term independent of $x$: $18 - 3r = 0 \implies 3r = 18 \implies r = 6$ So it is the 7th term ($T_7$).

$T_7 = \binom{9}{6} (-1)^6 \frac{3^{9 - 2(6)}}{2^{9 - 6}} = \binom{9}{3} (1) \frac{3^{-3}}{2^3}$ $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$ $T_7 = 84 \times \frac{1}{27} \times \frac{1}{8} = \frac{84}{216} = \frac{7}{18}$.