The term independent of x in `((1)/(2)x^(1//3)+x^(-1//5))^(8)` is

Options

A. 11
B. 10
C. 8
D. 7

Answer

To find the term independent of \( x \) in the expression \(\left(\frac{1}{2}x^{\frac{1}{3}} + x^{-\frac{1}{5}}\right)^{8}\), we will follow these steps:

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = \frac{1}{2}x^{\frac{1}{3}} \), \( b = x^{-\frac{1}{5}} \), and \( n = 8 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{8}{r} \left(\frac{1}{2}x^{\frac{1}{3}}\right)^{8-r} \left(x^{-\frac{1}{5}}\right)^{r} \]

### Step 2: Simplify the General Term
Now, we simplify the general term:
\[ T_{r+1} = \binom{8}{r} \left(\frac{1}{2}\right)^{8-r} x^{\frac{8-r}{3}} x^{-\frac{r}{5}} \]
Combining the powers of \( x \):
\[ T_{r+1} = \binom{8}{r} \left(\frac{1}{2}\right)^{8-r} x^{\frac{8-r}{3} - \frac{r}{5}} \]

### Step 3: Find the Power of \( x \)
To find the term that is independent of \( x \), we need the exponent of \( x \) to be zero:
\[ \frac{8-r}{3} - \frac{r}{5} = 0 \]

### Step 4: Solve for \( r \)
To solve for \( r \), we will first find a common denominator, which is 15:
\[ \frac{5(8-r)}{15} - \frac{3r}{15} = 0 \]
This simplifies to:
\[ 40 - 5r - 3r = 0 \]
\[ 40 - 8r = 0 \implies 8r = 40 \implies r = 5 \]

### Step 5: Substitute \( r \) Back into the General Term
Now that we have \( r = 5 \), we can find the term:
\[ T_{6} = \binom{8}{5} \left(\frac{1}{2}\right)^{8-5} x^{0} \]
Calculating \( \binom{8}{5} \):
\[ \binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
Now substituting back:
\[ T_{6} = 56 \left(\frac{1}{2}\right)^{3} = 56 \cdot \frac{1}{8} = 7 \]

### Final Answer
The term independent of \( x \) is \( 7 \).

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