The the term independent in the expansion of `[(t^(-1)-1)x+(t^(-1)+1)^(-1)x^(-1)]^8` is

Options

A. `56((1-t)/(1+t))^(3)`
B. `56((1+t)/(1-t))^(3)`
C. `70((1-t)/(1+t))^(4)`
D. `70((1+t)/(1-t))^(4)`

Answer

To find the term independent of \( x \) in the expansion of

\[ \left[(t^{-1}-1)x + (t^{-1}+1)^{-1}x^{-1}\right]^8, \]

we will follow these steps:

### Step 1: Simplify the expression

First, we need to simplify the expression inside the brackets. The term \((t^{-1}+1)^{-1}\) can be rewritten as:

\[ (t^{-1}+1)^{-1} = \frac{1}{t^{-1}+1} = \frac{t}{1+t}. \]

Thus, we can rewrite the expression as:

\[ \left[(t^{-1}-1)x + \frac{t}{1+t}x^{-1}\right]^8. \]

### Step 2: Identify the general term

Using the Binomial Theorem, the general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:

\[ T_r = \binom{n}{r} a^{n-r} b^r. \]

In our case, let:

\[ a = (t^{-1}-1)x \quad \text{and} \quad b = \frac{t}{1+t}x^{-1}. \]

Then the general term becomes:

\[ T_r = \binom{8}{r} \left((t^{-1}-1)x\right)^{8-r} \left(\frac{t}{1+t}x^{-1}\right)^r. \]

### Step 3: Expand the general term

Now, we can expand \( T_r \):

\[ T_r = \binom{8}{r} \left(t^{-1}-1\right)^{8-r} x^{8-r} \left(\frac{t}{1+t}\right)^r x^{-r}. \]

This simplifies to:

\[ T_r = \binom{8}{r} \left(t^{-1}-1\right)^{8-r} \left(\frac{t}{1+t}\right)^r x^{8-2r}. \]

### Step 4: Find the independent term of \( x \)

For the term to be independent of \( x \), we need \( 8 - 2r = 0 \). Solving this gives:

\[ 8 - 2r = 0 \implies 2r = 8 \implies r = 4. \]

### Step 5: Substitute \( r \) back into the general term

Now, substitute \( r = 4 \) back into the general term:

\[ T_4 = \binom{8}{4} \left(t^{-1}-1\right)^{4} \left(\frac{t}{1+t}\right)^{4}. \]

### Step 6: Calculate \( \binom{8}{4} \)

Calculating \( \binom{8}{4} \):

\[ \binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70. \]

### Step 7: Combine the terms

Thus, we have:

\[ T_4 = 70 \left(t^{-1}-1\right)^{4} \left(\frac{t}{1+t}\right)^{4}. \]

### Step 8: Final expression

Now, we can express \( T_4 \):

\[ T_4 = 70 \cdot (t^{-1}-1)^{4} \cdot \frac{t^4}{(1+t)^{4}}. \]

### Conclusion

The term independent of \( x \) in the expansion is:

\[ 70 \cdot (t^{-1}-1)^{4} \cdot \frac{t^4}{(1+t)^{4}}. \]