Find the term independent of x in the expansion of ` [1/2x^(1//3)+x^(-1//5)]^8`

Answer

To find the term independent of \( x \) in the expansion of \( \left( \frac{1}{2} x^{\frac{1}{3}} + x^{-\frac{1}{5}} \right)^8 \), 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, \( n = 8 \), \( a = \frac{1}{2} x^{\frac{1}{3}} \), and \( b = x^{-\frac{1}{5}} \).

2. **Write the general term for our expression**:
Substituting the values of \( a \) and \( b \):
\[ T_{r+1} = \binom{8}{r} \left( \frac{1}{2} x^{\frac{1}{3}} \right)^{8-r} \left( x^{-\frac{1}{5}} \right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{8}{r} \left( \frac{1}{2} \right)^{8-r} x^{\frac{8-r}{3}} x^{-\frac{r}{5}} \]
Combining the powers of \( x \):
\[ T_{r+1} = \binom{8}{r} \left( \frac{1}{2} \right)^{8-r} x^{\frac{8-r}{3} - \frac{r}{5}} \]

3. **Find the exponent of \( x \)**:
We need the exponent of \( x \) to be zero for the term to be independent of \( x \):
\[ \frac{8-r}{3} - \frac{r}{5} = 0 \]

4. **Solve for \( r \)**:
To solve the equation:
\[ \frac{8-r}{3} = \frac{r}{5} \]
Cross-multiplying gives:
\[ 5(8 - r) = 3r \]
Expanding and simplifying:
\[ 40 - 5r = 3r \implies 40 = 8r \implies r = 5 \]

5. **Determine the term independent of \( x \)**:
The term independent of \( x \) corresponds to \( r = 5 \). Thus, we find \( T_{6} \) (since \( T_{r+1} \)):
\[ T_{6} = \binom{8}{5} \left( \frac{1}{2} \right)^{8-5} x^{0} \]

6. **Calculate \( T_{6} \)**:
\[ T_{6} = \binom{8}{5} \left( \frac{1}{2} \right)^{3} \]
We know \( \binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \):
\[ T_{6} = 56 \cdot \frac{1}{8} = 7 \]

### Final Answer:
The term independent of \( x \) in the expansion is \( 7 \).