Find the term independent of x in `((3x^(2))/(2)-(1)/(3x))^(9)`

Answer

To find the term independent of \( x \) in the expression \(\left(\frac{3x^2}{2} - \frac{1}{3x}\right)^{9}\), we will use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the terms**:
We can identify \( a = \frac{3x^2}{2} \) and \( b = -\frac{1}{3x} \) with \( n = 9 \).

2. **Write the general term**:
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Substituting the values, we get:
\[ T_{r+1} = \binom{9}{r} \left(\frac{3x^2}{2}\right)^{9-r} \left(-\frac{1}{3x}\right)^r \]

3. **Simplify the general term**:
\[ T_{r+1} = \binom{9}{r} \left(\frac{3^{9-r} (x^2)^{9-r}}{2^{9-r}}\right) \left(-\frac{1}{3^r x^r}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{9}{r} \left(-1\right) \frac{3^{9-r}}{3^r} \frac{x^{2(9-r)}}{2^{9-r} x^r} \]
\[ = \binom{9}{r} \left(-1\right) \frac{3^{9}}{3^{2r} \cdot 2^{9-r}} x^{18 - 3r} \]

4. **Find the term independent of \( x \)**:
We want the term where the power of \( x \) is zero:
\[ 18 - 3r = 0 \]
Solving for \( r \):
\[ 3r = 18 \implies r = 6 \]

5. **Determine the specific term**:
Now, substituting \( r = 6 \) into the general term:
\[ T_{7} = \binom{9}{6} \left(-1\right) \frac{3^{9}}{3^{12} \cdot 2^{3}} x^{0} \]
\[ = \binom{9}{6} \left(-1\right) \frac{3^{9}}{27 \cdot 8} \]
\[ = \binom{9}{6} \left(-1\right) \frac{3^{6}}{8} \]
Since \(\binom{9}{6} = \binom{9}{3} = 84\):
\[ = 84 \cdot \left(-1\right) \cdot \frac{729}{8} \]
\[ = -\frac{84 \cdot 729}{8} \]

6. **Final result**:
The term independent of \( x \) is:
\[ -\frac{61416}{8} = -7677 \]