Factorise
` 27 -x^(3) y^(3) +6-2xy `

Answer

To factorise the expression \( 27 - x^3 y^3 + 6 - 2xy \), we can follow these steps:

### Step 1: Rearrange the expression
First, let's rearrange the expression to group similar terms:
\[ 27 + 6 - x^3 y^3 - 2xy \]
This simplifies to:
\[ 33 - x^3 y^3 - 2xy \]

### Step 2: Recognize the cubes
Notice that \( 27 \) can be expressed as \( 3^3 \) and \( x^3 y^3 \) can be expressed as \( (xy)^3 \). Thus, we can rewrite the expression:
\[ 3^3 - (xy)^3 + 33 - 2xy \]

### Step 3: Factor the difference of cubes
The expression \( 3^3 - (xy)^3 \) can be factored using the difference of cubes formula:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
Here, \( a = 3 \) and \( b = xy \). So we have:
\[ (3 - xy)(3^2 + 3(xy) + (xy)^2) \]
Calculating \( 3^2 + 3(xy) + (xy)^2 \):
\[ = 9 + 3xy + x^2y^2 \]
Thus, we can rewrite the expression as:
\[ (3 - xy)(9 + 3xy + x^2y^2) \]

### Step 4: Combine with the remaining terms
Now, we need to include the remaining \( +33 - 2xy \) into our factorization. We can rewrite \( 33 \) as \( 3 \times 11 \):
\[ 33 - 2xy = 3 \times 11 - 2xy \]
Now, we can factor out \( (3 - xy) \):
\[ (3 - xy)(9 + 3xy + x^2y^2) + 2(3 - xy) \]

### Step 5: Factor out the common term
Now we can factor out \( (3 - xy) \) from the entire expression:
\[ (3 - xy)(9 + 3xy + x^2y^2 + 2) \]
This simplifies to:
\[ (3 - xy)(x^2y^2 + 3xy + 11) \]

### Final Answer
Thus, the factorized form of the expression is:
\[ (3 - xy)(x^2y^2 + 3xy + 11) \]

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