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

Options

A. 9
B. 18
C. 48
D. 84

Answer

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

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

### Step 2: Simplify the general term
Now, we simplify the general term:
\[ T_r = \binom{9}{r} (x^{2(9-r)}) \left(-1\right)^r \left(\frac{1}{x^r}\right) \]
This simplifies to:
\[ T_r = \binom{9}{r} (-1)^r x^{18 - 2r - r} = \binom{9}{r} (-1)^r x^{18 - 3r} \]

### Step 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 \( r \):
\[ 3r = 18 \implies r = 6 \]

### Step 4: Substitute \( r \) back into the general term
Now, we substitute \( r = 6 \) back into the general term to find the value:
\[ T_6 = \binom{9}{6} (-1)^6 \]
Calculating \( \binom{9}{6} \):
\[ \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]
Thus, \( T_6 = 84 \).

### Conclusion
The term independent of \( x \) in the expansion of \( (x^2 - \frac{1}{x})^9 \) is:
\[ \boxed{84} \]