Find term which is independent of x in `(x^(2)-(1)/(x^(6)))^(16)`

Answer

To find the term that is independent of \( x \) in the expression \( \left( x^2 - \frac{1}{x^6} \right)^{16} \), we will use the binomial expansion.

### Step-by-Step Solution:

1. **Identify the Binomial Expansion**:
The expression can be expanded using the binomial theorem, which states that:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^r b^{n-r} \]
Here, let \( a = x^2 \) and \( b = -\frac{1}{x^6} \), and \( n = 16 \).

2. **Write the General Term**:
The general term \( T_r \) in the expansion is given by:
\[ T_r = \binom{16}{r} (x^2)^r \left(-\frac{1}{x^6}\right)^{16-r} \]
Simplifying this, we get:
\[ T_r = \binom{16}{r} x^{2r} \left(-1\right)^{16-r} \frac{1}{x^{6(16-r)}} \]
This can be rewritten as:
\[ T_r = \binom{16}{r} (-1)^{16-r} x^{2r - 6(16-r)} \]

3. **Combine the Exponents of \( x \)**:
The exponent of \( x \) in \( T_r \) is:
\[ 2r - 6(16 - r) = 2r - 96 + 6r = 8r - 96 \]
We need to find the value of \( r \) such that the exponent of \( x \) is zero:
\[ 8r - 96 = 0 \]

4. **Solve for \( r \)**:
Rearranging the equation gives:
\[ 8r = 96 \implies r = \frac{96}{8} = 12 \]

5. **Find the Independent Term**:
Now that we have \( r = 12 \), we can substitute this back into the general term:
\[ T_{12} = \binom{16}{12} (-1)^{16-12} x^{8(12) - 96} \]
Since \( 8(12) - 96 = 0 \), the term \( T_{12} \) is independent of \( x \).

6. **Calculate \( \binom{16}{12} \)**:
We can simplify \( \binom{16}{12} \) as follows:
\[ \binom{16}{12} = \binom{16}{4} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} \]
Calculating this gives:
\[ = \frac{16 \times 15 \times 14 \times 13}{24} \]
Simplifying further:
\[ = \frac{16 \times 15 \times 14 \times 13}{4 \times 6} = \frac{16 \times 15 \times 14 \times 13}{24} \]
Continuing the calculation:
\[ = 1820 \]

### Final Answer:
The term that is independent of \( x \) in the expansion of \( \left( x^2 - \frac{1}{x^6} \right)^{16} \) is \( 1820 \).