Factorise : `8x^(3)-27y^(3)-2x+3y` : The factors are

Options

A. `(2x-3y)(4x^(2)+xy+9y^(2)-1)`
B. `(2x-3y)(4x^(2)-6xy+9y^(2))`
C. `(2x-3y)(4x^(2)+3xy+9y^(2)-1)`
D. `(2x-3y)(4x^(2)+6xy+9y^(2)-1)`

Answer

To factorise the expression \( 8x^3 - 27y^3 - 2x + 3y \), we can follow these steps:

### Step 1: Rewrite the expression
We can express \( 8x^3 \) and \( 27y^3 \) as cubes:
\[ 8x^3 = (2x)^3 \quad \text{and} \quad 27y^3 = (3y)^3 \]
Thus, we rewrite the expression:
\[ (2x)^3 - (3y)^3 - 2x + 3y \]

### Step 2: Group the terms
Next, we can group the terms:
\[ ((2x)^3 - (3y)^3) + (-2x + 3y) \]

### Step 3: Apply the difference of cubes formula
Using the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \), where \( a = 2x \) and \( b = 3y \):
\[ (2x - 3y)((2x)^2 + (2x)(3y) + (3y)^2) \]
Calculating the second factor:
\[ (2x)^2 = 4x^2, \quad (2x)(3y) = 6xy, \quad (3y)^2 = 9y^2 \]
So, we have:
\[ (2x - 3y)(4x^2 + 6xy + 9y^2) \]

### Step 4: Combine with the remaining terms
Now, we need to combine this with the remaining terms \(-2x + 3y\):
\[ (2x - 3y)(4x^2 + 6xy + 9y^2) - (2x - 3y) \]

### Step 5: Factor out the common term
Notice that both terms have a common factor of \( (2x - 3y) \):
\[ (2x - 3y)((4x^2 + 6xy + 9y^2) - 1) \]

### Step 6: Final expression
Thus, the final factorised form of the expression is:
\[ (2x - 3y)(4x^2 + 6xy + 9y^2 - 1) \]

### Summary
The factors of the expression \( 8x^3 - 27y^3 - 2x + 3y \) are:
\[ (2x - 3y)(4x^2 + 6xy + 9y^2 - 1) \]