Term independent of x in `(x - 1/x)^4 (x + 1/x)^3` is

Options

A. 1
B. 2
C. 0
D. 4

Answer

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

### Step 1: Expand each binomial using the Binomial Theorem
We will first expand \( (x - \frac{1}{x})^4 \) and \( (x + \frac{1}{x})^3 \) separately.

**Expansion of \( (x - \frac{1}{x})^4 \)**:
Using the Binomial Theorem:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
Here, \( a = x \), \( b = -\frac{1}{x} \), and \( n = 4 \):
\[ (x - \frac{1}{x})^4 = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} \left(-\frac{1}{x}\right)^k = \sum_{k=0}^{4} \binom{4}{k} (-1)^k x^{4-2k} \]

This gives us the terms:
- For \( k = 0 \): \( \binom{4}{0} x^4 = x^4 \)
- For \( k = 1 \): \( \binom{4}{1} (-1) x^2 = -4x^2 \)
- For \( k = 2 \): \( \binom{4}{2} x^0 = 6 \)
- For \( k = 3 \): \( \binom{4}{3} (-1) \frac{1}{x^2} = -4 \frac{1}{x^2} \)
- For \( k = 4 \): \( \binom{4}{4} \frac{1}{x^4} = \frac{1}{x^4} \)

So, the expansion is:
\[ (x - \frac{1}{x})^4 = x^4 - 4x^2 + 6 - \frac{4}{x^2} + \frac{1}{x^4} \]

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

This gives us the terms:
- For \( j = 0 \): \( \binom{3}{0} x^3 = x^3 \)
- For \( j = 1 \): \( \binom{3}{1} x = 3x \)
- For \( j = 2 \): \( \binom{3}{2} \frac{1}{x} = 3 \frac{1}{x} \)
- For \( j = 3 \): \( \binom{3}{3} \frac{1}{x^3} = \frac{1}{x^3} \)

So, the expansion is:
\[ (x + \frac{1}{x})^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \]

### Step 2: Multiply the two expansions
Now, we need to multiply these two expansions together:
\[ (x - \frac{1}{x})^4 (x + \frac{1}{x})^3 \]

### Step 3: Identify the term independent of \( x \)
We need to find the combinations of terms from both expansions that result in \( x^0 \) (i.e., independent of \( x \)).

- From \( (x - \frac{1}{x})^4 \), the constant term is \( 6 \).
- From \( (x + \frac{1}{x})^3 \), the constant term is also \( 3 \).

So, the product of these constant terms gives us:
\[ 6 \cdot 3 = 18 \]

Next, we can find other combinations:
- \( (-4x^2) \) from the first expansion and \( \frac{3}{x} \) from the second gives \( -12 \).
- \( 6 \) from the first and \( 3x \) from the second gives \( 18 \).
- \( -4 \frac{1}{x^2} \) from the first and \( x^3 \) from the second gives \( -4 \).

Adding these contributions together:
\[ 18 - 12 + 18 - 4 = 20 \]

### Conclusion
The term independent of \( x \) in the expression \( (x - \frac{1}{x})^4 (x + \frac{1}{x})^3 \) is \( 20 \).