Find the term independent of `x` in `(2x^(2) - (1)/(x) )^(12)`.

Answer

To find the term independent of \( x \) in the expression \( (2x^2 - \frac{1}{x})^{12} \), we will use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the general term**:
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = 2x^2 \), \( b = -\frac{1}{x} \), and \( n = 12 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{12}{r} (2x^2)^{12-r} \left(-\frac{1}{x}\right)^r \]

2. **Simplify the general term**:
Expanding this, we get:
\[ T_{r+1} = \binom{12}{r} (2^{12-r} (x^2)^{12-r}) \left(-\frac{1}{x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{12}{r} 2^{12-r} (-1)^r x^{2(12-r) - r} \]
Therefore:
\[ T_{r+1} = \binom{12}{r} 2^{12-r} (-1)^r x^{24 - 3r} \]

3. **Find the term independent of \( x \)**:
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ 24 - 3r = 0 \]
Solving for \( r \):
\[ 3r = 24 \quad \Rightarrow \quad r = 8 \]

4. **Substitute \( r \) back into the general term**:
Now, we substitute \( r = 8 \) back into the expression for \( T_{r+1} \):
\[ T_{9} = \binom{12}{8} 2^{12-8} (-1)^8 \]
This simplifies to:
\[ T_{9} = \binom{12}{8} 2^4 \]

5. **Calculate \( \binom{12}{8} \)**:
We know that \( \binom{12}{8} = \binom{12}{4} \):
\[ \binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 \]

6. **Calculate \( 2^4 \)**:
\[ 2^4 = 16 \]

7. **Final calculation**:
Now, we can find the term independent of \( x \):
\[ T_{9} = 495 \times 16 = 7920 \]

### Conclusion:
The term independent of \( x \) in the expansion of \( (2x^2 - \frac{1}{x})^{12} \) is **7920**.