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

Options

A. 1
B. `-1`
C. 48
D. None

Answer

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

### Step-by-Step Solution:

1. **Identify the Binomial Expansion**:
The expression can be expanded using the Binomial Theorem:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
Here, let \( a = x^2 \) and \( b = -\frac{1}{x} \), and \( n = 9 \).

2. **Write the General Term**:
The general term \( T_r \) in the expansion is given by:
\[ T_r = \binom{9}{r} (x^2)^{9-r} \left(-\frac{1}{x}\right)^r \]
Simplifying this, we have:
\[ T_r = \binom{9}{r} (x^2)^{9-r} (-1)^r x^{-r} \]
\[ = \binom{9}{r} (-1)^r x^{2(9-r) - r} \]
\[ = \binom{9}{r} (-1)^r x^{18 - 2r - r} \]
\[ = \binom{9}{r} (-1)^r x^{18 - 3r} \]

3. **Find the Term Independent of \( x \)**:
To find the term independent of \( x \), we set the exponent of \( x \) to zero:
\[ 18 - 3r = 0 \]
Solving for \( r \):
\[ 3r = 18 \implies r = 6 \]

4. **Substitute \( r = 6 \) into the General Term**:
Now, we substitute \( r = 6 \) into the general term:
\[ T_6 = \binom{9}{6} (-1)^6 x^{18 - 3 \cdot 6} \]
\[ = \binom{9}{6} x^0 \]
Since \( x^0 = 1 \), we only need to calculate \( \binom{9}{6} \):
\[ \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \]

5. **Final Result**:
Therefore, the term independent of \( x \) in the expansion of \( (x^2 - \frac{1}{x})^9 \) is:
\[ \boxed{84} \]