Factorise :
` (a+ b) ^(2) - 5( a^(2) - b^(2) ) - 24 (a-b)^(2)`

Answer

To factorise the expression \( (a + b)^2 - 5(a^2 - b^2) - 24(a - b)^2 \), we will follow these steps:

### Step 1: Expand the terms
Start by expanding each part of the expression using the identities:
- \( (a + b)^2 = a^2 + b^2 + 2ab \)
- \( a^2 - b^2 = (a + b)(a - b) \)
- \( (a - b)^2 = a^2 - 2ab + b^2 \)

So, we rewrite the expression:
\[ (a + b)^2 - 5(a^2 - b^2) - 24(a - b)^2 \]
Substituting the identities:
\[ = (a^2 + b^2 + 2ab) - 5(a^2 - b^2) - 24(a^2 - 2ab + b^2) \]

### Step 2: Substitute and simplify
Substituting the identities into the expression:
\[ = a^2 + b^2 + 2ab - 5(a^2 - b^2) - 24(a^2 - 2ab + b^2) \]
Expanding the terms:
\[ = a^2 + b^2 + 2ab - 5a^2 + 5b^2 - 24a^2 + 48ab - 24b^2 \]

### Step 3: Combine like terms
Now, combine the like terms:
- For \( a^2 \): \( 1 - 5 - 24 = -28a^2 \)
- For \( b^2 \): \( 1 + 5 - 24 = -18b^2 \)
- For \( ab \): \( 2 + 48 = 50ab \)

Putting it all together, we have:
\[ -28a^2 - 18b^2 + 50ab \]

### Step 4: Factor out common terms
Now, we can factor out a common factor of -2:
\[ = -2(14a^2 + 9b^2 - 25ab) \]

### Step 5: Factor the quadratic expression
Next, we need to factor the quadratic \( 14a^2 - 25ab + 9b^2 \). We look for two numbers that multiply to \( 14 \times 9 = 126 \) and add to \( -25 \). The numbers are \( -18 \) and \( -7 \).

Rewriting the quadratic:
\[ = -2(14a^2 - 18ab - 7ab + 9b^2) \]
Grouping the terms:
\[ = -2[(14a^2 - 18ab) + (-7ab + 9b^2)] \]
Factoring by grouping:
\[ = -2[2a(7a - 9b) - b(7a - 9b)] \]
Factoring out \( (7a - 9b) \):
\[ = -2(7a - 9b)(2a - b) \]

### Final Answer
Thus, the factorised form of the expression is:
\[ -2(7a - 9b)(2a - b) \]