Fractorise:
`12(2x - 3y )^(2) -16 (3y - 2x )`

Answer

To factorise the expression \( 12(2x - 3y)^2 - 16(3y - 2x) \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ 12(2x - 3y)^2 - 16(3y - 2x) \]
Notice that \( 3y - 2x \) can be rewritten as \( -(2x - 3y) \). Thus, we can express the second term as:
\[ 12(2x - 3y)^2 + 16(2x - 3y) \]

### Step 2: Factor out the common term
Now, we can see that both terms contain \( (2x - 3y) \). Let’s factor \( (2x - 3y) \) out:
\[ = (2x - 3y) \left[ 12(2x - 3y) + 16 \right] \]

### Step 3: Simplify the expression inside the brackets
Next, we simplify the expression inside the brackets:
\[ 12(2x - 3y) + 16 \]
This can be simplified further:
\[ = 12(2x - 3y) + 16 = 12(2x - 3y) + 16 \]

### Step 4: Factor out the constant from the bracket
Now, we can factor out a common factor of 4 from the expression inside the brackets:
\[ = 4 \left[ 3(2x - 3y) + 4 \right] \]
Thus, we have:
\[ = (2x - 3y) \left[ 4(3(2x - 3y) + 4) \right] \]

### Step 5: Write the final factorised form
Putting it all together, we get:
\[ = 4(2x - 3y)(3(2x - 3y) + 4) \]

### Final Answer:
The factorised form of the expression is:
\[ 4(2x - 3y)(3(2x - 3y) + 4) \]