If `A =((x-1)/(x+1))` and `B =((x+1)/(x-1))`, then the value of `(A+B)^(2)` is:

Options

A. `(4x^(2) +8x^(2) + 4)/(x^(2) -2x-1)`
B. `(4x^(4) + 8x^(2)+4)/(x^(4) + 2x^(2)-1)`
C. `(4x^(4) + 8x^(2) +4)/(x^(4)- 2x^(2) +1)`
D. None of these

Answer

To solve the problem, we need to find the value of \((A + B)^2\) where \(A = \frac{x-1}{x+1}\) and \(B = \frac{x+1}{x-1}\).

### Step-by-Step Solution:

1. **Write down the expressions for A and B:**
\[ A = \frac{x - 1}{x + 1}, \quad B = \frac{x + 1}{x - 1} \]

2. **Find A + B:**
\[ A + B = \frac{x - 1}{x + 1} + \frac{x + 1}{x - 1} \]

3. **Find a common denominator:**
The common denominator for the two fractions is \((x + 1)(x - 1)\). Thus,
\[ A + B = \frac{(x - 1)(x - 1) + (x + 1)(x + 1)}{(x + 1)(x - 1)} \]

4. **Expand the numerators:**
\[ (x - 1)(x - 1) = (x - 1)^2 = x^2 - 2x + 1 \]
\[ (x + 1)(x + 1) = (x + 1)^2 = x^2 + 2x + 1 \]
Therefore,
\[ A + B = \frac{(x^2 - 2x + 1) + (x^2 + 2x + 1)}{(x + 1)(x - 1)} \]

5. **Combine the numerators:**
\[ A + B = \frac{2x^2 + 2}{(x + 1)(x - 1)} = \frac{2(x^2 + 1)}{(x + 1)(x - 1)} \]

6. **Now, square the result:**
\[ (A + B)^2 = \left(\frac{2(x^2 + 1)}{(x + 1)(x - 1)}\right)^2 = \frac{4(x^2 + 1)^2}{(x + 1)^2(x - 1)^2} \]

7. **Expand \((x^2 + 1)^2\):**
\[ (x^2 + 1)^2 = x^4 + 2x^2 + 1 \]

8. **Final expression:**
\[ (A + B)^2 = \frac{4(x^4 + 2x^2 + 1)}{(x + 1)^2(x - 1)^2} \]

### Final Answer:
\[ (A + B)^2 = \frac{4(x^4 + 2x^2 + 1)}{(x + 1)^2(x - 1)^2} \]