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

Options

A. `28//81`
B. `28//243`
C. `-28//243`
D. `-28//81`

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 \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x^2 \), \( b = -\frac{1}{3x} \), and \( n = 9 \). Therefore, the general term becomes:
\[ T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(-\frac{1}{3x}\right)^r \]

### Step 2: Simplify the General Term
Now, we simplify the general term:
\[ T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(-\frac{1}{3}\right)^r (x^{-1})^r \]
This can be rewritten as:
\[ T_{r+1} = \binom{9}{r} (-1)^r \frac{(x^2)^{9-r}}{3^r x^r} = \binom{9}{r} (-1)^r \frac{x^{18 - 2r}}{3^r} \]

### Step 3: Find the Condition for Independence from \( x \)
To find the term independent of \( x \), we need the exponent of \( x \) to be zero:
\[ 18 - 2r - r = 0 \]
This simplifies to:
\[ 18 - 3r = 0 \implies 3r = 18 \implies r = 6 \]

### Step 4: Substitute \( r \) Back into the General Term
Now, we substitute \( r = 6 \) back into the general term to find \( T_{7} \):
\[ T_{7} = \binom{9}{6} (-1)^6 \frac{1}{3^6} = \binom{9}{6} \frac{1}{729} \]
Since \( \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \), we have:
\[ T_{7} = 84 \cdot \frac{1}{729} = \frac{84}{729} \]

### Step 5: Simplify the Result
Now we simplify \( \frac{84}{729} \):
\[ \frac{84}{729} = \frac{28}{243} \]

### Final Answer
Thus, the term independent of \( x \) in the expansion of \( (x^2 - \frac{1}{3x})^9 \) is:
\[ \frac{28}{243} \]