The term independent of x in the expansion of `(x^2-(1)/(x))^6` is

Options

A. `-12`
B. 15
C. 24
D. `-15`

Answer

To find the term independent of \( x \) in the expansion of \( (x^2 - \frac{1}{x})^6 \), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x^2 \), \( b = -\frac{1}{x} \), and \( n = 6 \).

### Step 2: Write the general term for our specific case
Thus, the general term becomes:
\[ T_{r+1} = \binom{6}{r} (x^2)^{6-r} \left(-\frac{1}{x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-1)^r x^{-r} \]
\[ = \binom{6}{r} (-1)^r x^{12 - 2r - r} \]
\[ = \binom{6}{r} (-1)^r x^{12 - 3r} \]

### Step 3: Set the exponent of \( x \) to zero for the term independent of \( x \)
To find the term independent of \( x \), we need:
\[ 12 - 3r = 0 \]
Solving for \( r \):
\[ 3r = 12 \implies r = 4 \]

### Step 4: Substitute \( r \) back into the general term
Now, substitute \( r = 4 \) back into the expression for the general term:
\[ T_{4+1} = \binom{6}{4} (-1)^4 x^{12 - 3 \cdot 4} \]
\[ = \binom{6}{4} (1) x^0 \]
\[ = \binom{6}{4} \]

### Step 5: Calculate \( \binom{6}{4} \)
Using the formula for combinations:
\[ \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

### Conclusion
Thus, the term independent of \( x \) in the expansion of \( (x^2 - \frac{1}{x})^6 \) is \( 15 \).