In the expression of `(x^(4/5) + x^(-1/5))^(n)`, the coefficient of the `8^(th)` and `19^(th)` terms are equal. The term independent of x is given by

Options

A. `.^(27)C_(21)`
B. `.^(25)C_(20)`
C. `.^(25)C_(21)`
D. `.^(27)C_(22)`

Answer

To solve the problem, we need to find the term independent of \( x \) in the expression \( (x^{4/5} + x^{-1/5})^n \) given that the coefficients of the 8th and 19th terms are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_k = \binom{n}{k-1} a^{n-(k-1)} b^{k-1} \]
For our expression, let \( a = x^{4/5} \) and \( b = x^{-1/5} \). Thus, the general term becomes:
\[ T_k = \binom{n}{k-1} (x^{4/5})^{n-(k-1)} (x^{-1/5})^{k-1} \]
Simplifying this, we get:
\[ T_k = \binom{n}{k-1} x^{\frac{4(n-k+1)}{5} - \frac{k-1}{5}} = \binom{n}{k-1} x^{\frac{4n - 4k + 4 - k + 1}{5}} = \binom{n}{k-1} x^{\frac{4n - 5k + 5}{5}} \]

2. **Find the 8th and 19th Terms**:
- The 8th term corresponds to \( k = 8 \):
\[ T_8 = \binom{n}{7} x^{\frac{4n - 5 \cdot 8 + 5}{5}} = \binom{n}{7} x^{\frac{4n - 40 + 5}{5}} = \binom{n}{7} x^{\frac{4n - 35}{5}} \]
- The 19th term corresponds to \( k = 19 \):
\[ T_{19} = \binom{n}{18} x^{\frac{4n - 5 \cdot 19 + 5}{5}} = \binom{n}{18} x^{\frac{4n - 95 + 5}{5}} = \binom{n}{18} x^{\frac{4n - 90}{5}} \]

3. **Set the Coefficients Equal**:
Since the coefficients of the 8th and 19th terms are equal:
\[ \binom{n}{7} = \binom{n}{18} \]
Using the property of binomial coefficients, we know:
\[ \binom{n}{r} = \binom{n}{n-r} \]
This gives us:
\[ 7 + 18 = n \implies n = 25 \]

4. **Find the Term Independent of \( x \)**:
We need to find the term where the exponent of \( x \) is zero:
\[ \frac{4n - 5k + 5}{5} = 0 \]
Substituting \( n = 25 \):
\[ 4(25) - 5k + 5 = 0 \implies 100 - 5k + 5 = 0 \implies 105 - 5k = 0 \implies 5k = 105 \implies k = 21 \]

5. **Calculate the Independent Term**:
The independent term corresponds to \( k = 21 \):
\[ T_{21} = \binom{25}{20} (x^{4/5})^{25-20} (x^{-1/5})^{20} = \binom{25}{20} (x^{4/5})^{5} (x^{-1/5})^{20} \]
Simplifying:
\[ T_{21} = \binom{25}{20} x^{4 - 4} = \binom{25}{20} = \binom{25}{5} \]

### Final Answer:
The term independent of \( x \) is \( \binom{25}{5} \).