Factorise :
`1-2a - 2b - 3(a+b)^(2)`

Answer

To factorise the expression \(1 - 2a - 2b - 3(a + b)^2\), we will follow these steps:

### Step 1: Rewrite the expression
Start with the given expression:
\[ 1 - 2a - 2b - 3(a + b)^2 \]

### Step 2: Expand the square term
Expand the term \(3(a + b)^2\):
\[ 3(a + b)^2 = 3(a^2 + 2ab + b^2) \]
So, we rewrite the expression as:
\[ 1 - 2a - 2b - 3(a^2 + 2ab + b^2) \]

### Step 3: Combine like terms
Now, combine the terms:
\[ 1 - 2a - 2b - 3a^2 - 6ab - 3b^2 \]
This simplifies to:
\[ 1 - 3a^2 - 2a - 6ab - 3b^2 - 2b \]

### Step 4: Group the terms
Group the terms in a way that we can factor them:
\[ 1 - (3a^2 + 6ab + 3b^2 + 2a + 2b) \]

### Step 5: Substitute \(x = a + b\)
Let \(x = a + b\). Then, we can rewrite \(2a + 2b\) as \(2x\):
\[ 1 - 3(a + b)^2 - 2(a + b) = 1 - 3x^2 - 2x \]

### Step 6: Rearranging the expression
Now, we have:
\[ 1 - 3x^2 - 2x \]

### Step 7: Factor the quadratic expression
This expression can be rearranged as:
\[ -3x^2 - 2x + 1 \]
To factor this, we can look for two numbers that multiply to \(-3\) (the product of the coefficient of \(x^2\) and the constant term) and add to \(-2\) (the coefficient of \(x\)). The numbers are \(-3\) and \(1\).

So we can write:
\[ -3x^2 - 3x + x + 1 \]

### Step 8: Factor by grouping
Group the terms:
\[ (-3x^2 - 3x) + (x + 1) \]
Factoring out common terms gives:
\[ -3x(x + 1) + 1(x + 1) \]
Now factor out \((x + 1)\):
\[ (1 - 3x)(x + 1) \]

### Step 9: Substitute back for \(x\)
Now substitute back \(x = a + b\):
\[ (1 - 3(a + b))(a + b + 1) \]

### Final Answer
Thus, the factorised form of the expression \(1 - 2a - 2b - 3(a + b)^2\) is:
\[ (1 - 3(a + b))(a + b + 1) \]

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