Factorise :
`(1-x^(2))(1-y^(2)) + 4xy`

Answer

To factorise the expression \((1 - x^2)(1 - y^2) + 4xy\), we can follow these steps:

### Step 1: Expand the expression
We start by expanding the first part of the expression:

\[ (1 - x^2)(1 - y^2) = 1 - y^2 - x^2 + x^2y^2 \]

Now, we can rewrite the entire expression:

\[ 1 - y^2 - x^2 + x^2y^2 + 4xy \]

### Step 2: Rearrange the terms
Next, we rearrange the terms for clarity:

\[ 1 - x^2 - y^2 + x^2y^2 + 4xy \]

### Step 3: Group the terms
Now, we group the terms that can be combined:

\[ 1 + (x^2y^2 + 4xy) - (x^2 + y^2) \]

### Step 4: Factor the quadratic expression
We can factor the expression \(x^2y^2 + 4xy\) as follows:

\[ x^2y^2 + 4xy = (xy)^2 + 4xy = xy(xy + 4) \]

### Step 5: Combine the expression
Now, we can rewrite the expression:

\[ 1 + xy(xy + 4) - (x^2 + y^2) \]

### Step 6: Recognize the perfect square
Notice that \(x^2 + y^2 - 2xy\) can be rewritten as \((x - y)^2\). Thus, we can express the entire expression as:

\[ 1 + (xy + 2)^2 - (x - y)^2 \]

### Step 7: Apply the difference of squares
Now we can apply the difference of squares formula:

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

Let \(a = xy + 2\) and \(b = x - y\):

\[ (1 + (xy + 2) + (x - y))(1 + (xy + 2) - (x - y)) \]

### Final Expression
Thus, the factorised form of the original expression is:

\[ (1 + xy + 2 + x - y)(1 + xy + 2 - (x - y)) \]

This simplifies to:

\[ (3 + xy + x - y)(3 + xy - x + y) \]