Factorise: `27 a ^(3) - b ^(3) - 1 - 9 ab`

Options

A. `(3 a + b -1) (9a ^(2) + b ^(2) + 1 - 3 ab + 3 a - b) `
B. `(3a - b - 1) ( 9a ^(2) - b ^(2) + 1 + 3 ab + 3a - b )`
C. `(3a + b + 1) ( 9a ^(2) + 1 + 3 ab + 3a - b)`
D. `(3a-b -1) ( 9a ^(2) + b ^(2) + 1 + 3 ab + 3a - b)`

Answer

To factorize the expression \(27a^3 - b^3 - 1 - 9ab\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ 27a^3 - b^3 - 1 - 9ab \]

### Step 2: Group the terms
We can rearrange the terms to group them in a way that highlights their structure:
\[ 27a^3 - 9ab - b^3 - 1 \]

### Step 3: Identify the cubes
Notice that \(27a^3\) is \((3a)^3\), \(b^3\) is \(b^3\), and \(1\) is \(1^3\). We can rewrite the expression as:
\[ (3a)^3 - b^3 - 1^3 - 9ab \]

### Step 4: Recognize the identity
We can apply the identity for the sum of cubes:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) \]
In our case, let:
- \(x = 3a\)
- \(y = -b\)
- \(z = -1\)

### Step 5: Substitute into the identity
Now we substitute \(x\), \(y\), and \(z\) into the identity:
- \(x + y + z = 3a - b - 1\)
- \(x^2 = (3a)^2 = 9a^2\)
- \(y^2 = (-b)^2 = b^2\)
- \(z^2 = (-1)^2 = 1\)
- \(xy = (3a)(-b) = -3ab\)
- \(xz = (3a)(-1) = -3a\)
- \(yz = (-b)(-1) = b\)

### Step 6: Form the second factor
Now we can form the second factor:
\[ x^2 + y^2 + z^2 - xy - xz - yz = 9a^2 + b^2 + 1 - (-3ab) - (-3a) - b \]
This simplifies to:
\[ 9a^2 + b^2 + 1 + 3ab + 3a - b \]

### Step 7: Write the final factorization
Thus, the factorization of the expression is:
\[ (3a - b - 1)(9a^2 + b^2 + 1 + 3ab + 3a - b) \]

### Final Answer
The final factorized form of \(27a^3 - b^3 - 1 - 9ab\) is:
\[ (3a - b - 1)(9a^2 + b^2 + 1 + 3ab + 3a - b) \]