Solve each of the following quadratic equations:
`4x^(2)-4a^(2)x+(a^(4)-b^(4))=0`

Explanation

To solve the quadratic equation \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \), we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \( ax^2 + bx + c = 0 \).
Here,
- \( a = 4 \)
- \( b = -4a^2 \)
- \( c = a^4 - b^4 \)

### Step 2: Factor the quadratic equation
We can rewrite \( c \) using the difference of squares:
\[ a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) \]
Thus, the equation becomes:
\[ 4x^2 - 4a^2x + (a^2 - b^2)(a^2 + b^2) = 0 \]

### Step 3: Rewrite the middle term
We need to factor the quadratic. We can break the middle term \( -4a^2x \) into two parts that will help us factor the equation:
\[ -4a^2x = -2a^2x - 2a^2x \]
Now we can rewrite the equation:
\[ 4x^2 - 2a^2x - 2a^2x + (a^2 - b^2)(a^2 + b^2) = 0 \]

### Step 4: Group the terms
Now, we can group the terms:
\[ (4x^2 - 2a^2x) + (-2a^2x + (a^2 - b^2)(a^2 + b^2)) = 0 \]
Factoring out common terms:
\[ 2x(2x - a^2) - 2a^2(2x - (a^2 - b^2)) = 0 \]

### Step 5: Factor further
Now we can factor out \( (2x - (a^2 - b^2)) \):
\[ (2x - (a^2 - b^2))(2x - (a^2 + b^2)) = 0 \]

### Step 6: Set each factor to zero
Now we can set each factor equal to zero to find the roots:
1. \( 2x - (a^2 - b^2) = 0 \)
2. \( 2x - (a^2 + b^2) = 0 \)

### Step 7: Solve for \( x \)
From the first equation:
\[ 2x = a^2 - b^2 \implies x = \frac{a^2 - b^2}{2} \]

From the second equation:
\[ 2x = a^2 + b^2 \implies x = \frac{a^2 + b^2}{2} \]

### Final Solution
The roots of the quadratic equation \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \) are:
\[ x = \frac{a^2 - b^2}{2} \quad \text{and} \quad x = \frac{a^2 + b^2}{2} \]