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

Answer

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

### Step 1: Expand the Expression
We start by expanding the product \((a^{2} - 1)(b^{2} - 1)\).

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

Now, we can rewrite the original expression:

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

### Step 2: Rearrange the Terms
Next, we rearrange the terms to group similar ones:

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

### Step 3: Recognize a Perfect Square
Now, we can recognize that \(a^{2}b^{2} + 4ab + 1\) can be expressed as a perfect square. We can rewrite it as:

\[ (ab)^{2} + 2 \cdot ab \cdot 2 + 1^{2} = (ab + 2)^{2} \]

Thus, we have:

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

### Step 4: Apply the Difference of Squares Formula
Now we can apply the difference of squares formula, which states that \(x^{2} - y^{2} = (x - y)(x + y)\).

Let \(x = ab + 2\) and \(y = a + b\). Therefore, we can write:

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

### Step 5: Simplify the Factors
Now we simplify each factor:

1. For the first factor:
\[ ab + 2 - a - b = ab - a - b + 2 \]
2. For the second factor:
\[ ab + 2 + a + b = ab + a + b + 2 \]

### Final Factorised Form
Thus, the final factorised form of the expression is:

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