Factorise :
`a^(2) + b^(2) - c^(2) - d^(2) + 2ab - 2cd`

Answer

To factorise the expression \( a^2 + b^2 - c^2 - d^2 + 2ab - 2cd \), we can follow these steps:

### Step 1: Group the terms
We start by rearranging and grouping the terms in the expression:
\[ a^2 + b^2 + 2ab - c^2 - d^2 - 2cd \]

### Step 2: Recognize the perfect square
The first three terms \( a^2 + b^2 + 2ab \) can be recognized as a perfect square:
\[ a^2 + b^2 + 2ab = (a + b)^2 \]

Similarly, we can factor the last three terms:
\[ -c^2 - d^2 - 2cd = -(c^2 + d^2 + 2cd) = -(c + d)^2 \]

### Step 3: Rewrite the expression
Now, we can rewrite the expression using the perfect squares we identified:
\[ (a + b)^2 - (c + d)^2 \]

### Step 4: Apply the difference of squares formula
Now we can apply the difference of squares formula, which states that \( x^2 - y^2 = (x + y)(x - y) \). Here, let \( x = a + b \) and \( y = c + d \):
\[ (a + b)^2 - (c + d)^2 = (a + b + c + d)(a + b - (c + d)) \]

### Step 5: Simplify the expression
Now, we simplify the second term:
\[ a + b - (c + d) = a + b - c - d \]

### Final Factorized Form
Thus, the final factorized form of the expression is:
\[ (a + b + c + d)(a + b - c - d) \]