Factorise
` (1-a^(2))(1-b^(2)) + 4ab `

Answer

To factorize the expression \( (1 - a^2)(1 - b^2) + 4ab \), we will follow these steps:

### Step 1: Expand the expression
We start by expanding \( (1 - a^2)(1 - b^2) \).

\[ (1 - a^2)(1 - b^2) = 1 \cdot 1 - 1 \cdot b^2 - a^2 \cdot 1 + a^2 \cdot b^2 = 1 - b^2 - a^2 + a^2b^2 \]

Now, we add \( 4ab \) to this expression:

\[ 1 - b^2 - a^2 + a^2b^2 + 4ab \]

### Step 2: Rearrange the terms
Rearranging the terms gives us:

\[ 1 + a^2b^2 - a^2 - b^2 + 4ab \]

### Step 3: Group the terms
Next, we can group the terms related to \( a \) and \( b \):

\[ 1 + (a^2b^2 - a^2 - b^2 + 4ab) \]

### Step 4: Factor the quadratic expression
Now we can rewrite \( a^2b^2 - a^2 - b^2 + 4ab \) as:

\[ a^2b^2 + 4ab - a^2 - b^2 \]

We can factor this as:

\[ (a^2 + 4ab + b^2) - (a^2 + b^2) \]

### Step 5: Recognize the perfect square
Notice that \( a^2 + 4ab + b^2 \) can be rewritten as:

\[ (a + 2b)^2 \]

So we have:

\[ 1 + (a + 2b)^2 - (a^2 + b^2) \]

### Step 6: Apply the difference of squares
Now we can apply the difference of squares:

\[ (1 + (ab)^2) - (a - b)^2 \]

Using the identity \( x^2 - y^2 = (x + y)(x - y) \):

Let \( x = (1 + ab) \) and \( y = (a - b) \):

\[ (1 + ab + (a - b))(1 + ab - (a - b)) \]

### Final Result
Thus, the factorized form of the expression \( (1 - a^2)(1 - b^2) + 4ab \) is:

\[ (1 + ab + a - b)(1 + ab - a + b) \]