The term independent of `x` in the expansion of `(x-1/x)^4(x+1/x)^3` is: `-3` b. `0` c. `1` d. `3`

Options

A. -3
B. 0
C. 1
D. 3

Answer

To find the term independent of \( x \) in the expansion of \( (x - \frac{1}{x})^4 (x + \frac{1}{x})^3 \), we can follow these steps:

### Step 1: Expand Each Factor
First, we'll expand \( (x - \frac{1}{x})^4 \) and \( (x + \frac{1}{x})^3 \) separately using the binomial theorem.

1. **Expansion of \( (x - \frac{1}{x})^4 \)**:
\[ (x - \frac{1}{x})^4 = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} \left(-\frac{1}{x}\right)^k = \sum_{k=0}^{4} \binom{4}{k} (-1)^k x^{4-2k} \]

2. **Expansion of \( (x + \frac{1}{x})^3 \)**:
\[ (x + \frac{1}{x})^3 = \sum_{j=0}^{3} \binom{3}{j} x^{3-j} \left(\frac{1}{x}\right)^j = \sum_{j=0}^{3} \binom{3}{j} x^{3-2j} \]

### Step 2: Combine the Expansions
Now we need to combine these two expansions:
\[ (x - \frac{1}{x})^4 (x + \frac{1}{x})^3 = \left( \sum_{k=0}^{4} \binom{4}{k} (-1)^k x^{4-2k} \right) \left( \sum_{j=0}^{3} \binom{3}{j} x^{3-2j} \right) \]

### Step 3: Find the Term Independent of \( x \)
We need to find the terms where the total exponent of \( x \) is zero, i.e., \( 4 - 2k + 3 - 2j = 0 \) or \( 7 - 2k - 2j = 0 \).

This simplifies to:
\[ 2k + 2j = 7 \quad \Rightarrow \quad k + j = \frac{7}{2} \]
Since \( k \) and \( j \) must be integers, there are no integer solutions to this equation. Thus, there are no terms in the expansion that are independent of \( x \).

### Conclusion
The term independent of \( x \) in the expansion is \( 0 \).

### Final Answer
The answer is \( \boxed{0} \).