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

Options

A. `(28)/(91)`
B. `(28)/(243)`
C. `(27)/(243)`
D. `-(28)/(243)`

Answer

To find the term independent of \( x \) in the expansion of \( (x^2 - \frac{1}{3x})^9 \), we will follow these steps:

### Step 1: Identify the General Term
The general term \( T_r \) in the binomial expansion of \( (a + b)^n \) is given by:

\[ T_r = \binom{n}{r} a^{n-r} b^r \]

In our case, \( a = x^2 \) and \( b = -\frac{1}{3x} \), and \( n = 9 \). Thus, the general term becomes:

\[ T_r = \binom{9}{r} (x^2)^{9-r} \left(-\frac{1}{3x}\right)^r \]

### Step 2: Simplify the General Term
Now we simplify \( T_r \):

\[ T_r = \binom{9}{r} (x^2)^{9-r} \left(-\frac{1}{3}\right)^r (x^{-1})^r \]

This simplifies to:

\[ T_r = \binom{9}{r} (-1)^r \frac{1}{3^r} x^{2(9-r) - r} \]

### Step 3: Find the Power of \( x \)
Now we need to find the power of \( x \):

\[ \text{Power of } x = 2(9 - r) - r = 18 - 2r - r = 18 - 3r \]

### Step 4: Set the Power of \( x \) to Zero
To find the term independent of \( x \), we set the power of \( x \) to zero:

\[ 18 - 3r = 0 \]

### Step 5: Solve for \( r \)
Solving for \( r \):

\[ 3r = 18 \implies r = 6 \]

### Step 6: Substitute \( r \) Back into the General Term
Now, we substitute \( r = 6 \) back into the general term \( T_r \):

\[ T_6 = \binom{9}{6} (-1)^6 \frac{1}{3^6} x^{18 - 3 \cdot 6} \]

Since \( (-1)^6 = 1 \) and \( x^{18 - 18} = x^0 \):

\[ T_6 = \binom{9}{6} \frac{1}{3^6} \]

### Step 7: Calculate \( \binom{9}{6} \)
Using the property \( \binom{n}{r} = \binom{n}{n-r} \):

\[ \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

### Step 8: Calculate the Independent Term
Now we calculate the independent term:

\[ T_6 = 84 \cdot \frac{1}{729} = \frac{84}{729} \]

### Final Result
Thus, the term independent of \( x \) in the expansion of \( (x^2 - \frac{1}{3x})^9 \) is:

\[ \frac{28}{243} \]