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

Answer

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

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( n = 9 \), \( a = 2x^2 \), and \( b = -\frac{3}{x} \). Thus, the general term becomes:
\[ T_{r+1} = \binom{9}{r} (2x^2)^{9-r} \left(-\frac{3}{x}\right)^r \]

2. **Simplify the General Term**:
Expanding this, we get:
\[ T_{r+1} = \binom{9}{r} (2^{9-r} (x^2)^{9-r}) \left(-3^r \cdot x^{-r}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{9}{r} (-1)^r 2^{9-r} 3^r x^{2(9-r) - r} \]
Which further simplifies to:
\[ T_{r+1} = \binom{9}{r} (-1)^r 2^{9-r} 3^r x^{18 - 3r} \]

3. **Set the Power of \( x \) to Zero**:
To find the term independent of \( x \), we need the exponent of \( x \) to be zero:
\[ 18 - 3r = 0 \]
Solving for \( r \):
\[ 3r = 18 \quad \Rightarrow \quad r = 6 \]

4. **Substitute \( r \) Back into the General Term**:
Now, substituting \( r = 6 \) into the general term:
\[ T_{7} = \binom{9}{6} (-1)^6 2^{9-6} 3^6 x^{18 - 3 \cdot 6} \]
This simplifies to:
\[ T_{7} = \binom{9}{6} 2^3 3^6 \]

5. **Calculate the Binomial Coefficient**:
We know that \( \binom{9}{6} = \binom{9}{3} \):
\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

6. **Calculate the Final Value**:
Now, substituting the values:
\[ T_{7} = 84 \cdot 2^3 \cdot 3^6 \]
\[ = 84 \cdot 8 \cdot 729 \]
\[ = 672 \cdot 729 \]
\[ = 489888 \]

Thus, the term independent of \( x \) in the expansion of \( (2x^2 - \frac{3}{x})^9 \) is \( 489888 \).