If `(x^(4) - 2x^(2)y^(2) + y^(4))^(a-1) =(x-y)^(2a) (x+y)^(-2)`, then the value of a is:

Options

A. `(x^(2) - y^(2))`
B. `(x-y)/(x+y)`
C. `(log (x+y))/(log(x-y))`
D. `(log(x-y))/(log(x+y))`

Answer

To solve the equation \((x^{4} - 2x^{2}y^{2} + y^{4})^{(a-1)} = (x-y)^{2a} (x+y)^{-2}\), we will first simplify the left-hand side and then equate the exponents.

### Step 1: Simplify the left-hand side
The expression \(x^{4} - 2x^{2}y^{2} + y^{4}\) can be recognized as a perfect square. It can be factored as:
\[ x^{4} - 2x^{2}y^{2} + y^{4} = (x^{2} - y^{2})^{2} \]
Thus, we can rewrite the left-hand side:
\[ (x^{4} - 2x^{2}y^{2} + y^{4})^{(a-1)} = ((x^{2} - y^{2})^{2})^{(a-1)} = (x^{2} - y^{2})^{2(a-1)} \]

### Step 2: Rewrite the right-hand side
The right-hand side is already in a suitable form:
\[ (x-y)^{2a} (x+y)^{-2} \]

### Step 3: Express \(x^{2} - y^{2}\) in terms of \(x-y\) and \(x+y\)
Recall the identity:
\[ x^{2} - y^{2} = (x-y)(x+y) \]
Thus, we can express \((x^{2} - y^{2})^{2(a-1)}\) as:
\[ ((x-y)(x+y))^{2(a-1)} = (x-y)^{2(a-1)}(x+y)^{2(a-1)} \]

### Step 4: Equate the two sides
Now we have:
\[ (x-y)^{2(a-1)}(x+y)^{2(a-1)} = (x-y)^{2a}(x+y)^{-2} \]
We can equate the coefficients of \((x-y)\) and \((x+y)\) on both sides.

### Step 5: Equate the powers of \((x-y)\)
From \((x-y)\):
\[ 2(a-1) = 2a \implies 2a - 2 = 2a \implies -2 = 0 \]
This equation does not provide any new information.

### Step 6: Equate the powers of \((x+y)\)
From \((x+y)\):
\[ 2(a-1) = -2 \implies 2a - 2 = -2 \implies 2a = 0 \implies a = 0 \]

### Conclusion
Thus, the value of \(a\) is:
\[ \boxed{0} \]