Find the term independent of x (constant term) in the following expansion:
`(i) (x^(2)-(1)/(3x))^(9) " "(ii) (x-(1)/(x))^(10)`

Explanation

To find the term independent of \( x \) (constant term) in the given expansions, we will use the Binomial Theorem. Let's solve each part step by step.

### Part (i): \( (x^2 - \frac{1}{3x})^9 \)

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = x^2 \), \( b = -\frac{1}{3x} \), and \( n = 9 \).

Thus, the general term becomes:
\[ T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(-\frac{1}{3x}\right)^r \]

2. **Simplify the General Term**:
\[ T_{r+1} = \binom{9}{r} (x^{2(9-r)}) \left(-\frac{1}{3}\right)^r (x^{-r}) \]
\[ = \binom{9}{r} (-1)^r \frac{1}{3^r} x^{18 - 3r} \]

3. **Find the Term Independent of \( x \)**:
We want the exponent of \( x \) to be zero:
\[ 18 - 3r = 0 \]
Solving for \( r \):
\[ 3r = 18 \implies r = 6 \]

4. **Substitute \( r \) Back into the General Term**:
Now, substitute \( r = 6 \) into the general term:
\[ T_{7} = \binom{9}{6} (-1)^6 \frac{1}{3^6} x^{18 - 18} = \binom{9}{6} \frac{1}{729} \]

5. **Calculate \( \binom{9}{6} \)**:
\[ \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]
Therefore, the constant term is:
\[ T_{7} = 84 \times \frac{1}{729} = \frac{84}{729} = \frac{28}{243} \]

### Part (ii): \( (x - \frac{1}{x})^{10} \)

1. **Identify the General Term**:
The general term \( T_{r+1} \) in this case is:
\[ T_{r+1} = \binom{10}{r} x^{10-r} \left(-\frac{1}{x}\right)^r \]

2. **Simplify the General Term**:
\[ T_{r+1} = \binom{10}{r} x^{10-r} (-1)^r x^{-r} = \binom{10}{r} (-1)^r x^{10 - 2r} \]

3. **Find the Term Independent of \( x \)**:
Set the exponent of \( x \) to zero:
\[ 10 - 2r = 0 \]
Solving for \( r \):
\[ 2r = 10 \implies r = 5 \]

4. **Substitute \( r \) Back into the General Term**:
Substitute \( r = 5 \):
\[ T_{6} = \binom{10}{5} (-1)^5 x^{10 - 10} = \binom{10}{5} (-1) \]

5. **Calculate \( \binom{10}{5} \)**:
\[ \binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]
Therefore, the constant term is:
\[ T_{6} = -252 \]

### Final Answers:
- For part (i): The term independent of \( x \) is \( \frac{28}{243} \).
- For part (ii): The term independent of \( x \) is \( -252 \).