Factorise `: 16 (2x - y) ^(2) - 24 (4x ^(2) - y ^(2) ) + 9 ( 2x + y) ^(2)`

Answer

To factorize the expression \( 16(2x - y)^2 - 24(4x^2 - y^2) + 9(2x + y)^2 \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ 16(2x - y)^2 - 24(4x^2 - y^2) + 9(2x + y)^2 \]

### Step 2: Recognize the structure
Notice that \( 4x^2 - y^2 \) can be factored using the difference of squares:
\[ 4x^2 - y^2 = (2x - y)(2x + y) \]
Thus, we can rewrite the expression as:
\[ 16(2x - y)^2 - 24(2x - y)(2x + y) + 9(2x + y)^2 \]

### Step 3: Let \( a = 2x - y \) and \( b = 2x + y \)
Now, we can substitute \( a \) and \( b \) into our expression:
\[ 16a^2 - 24ab + 9b^2 \]

### Step 4: Recognize it as a quadratic in terms of \( a \)
This expression can be treated as a quadratic in \( a \):
\[ 16a^2 - 24ab + 9b^2 \]
We can factor this quadratic expression.

### Step 5: Factor the quadratic
To factor \( 16a^2 - 24ab + 9b^2 \), we look for two numbers that multiply to \( 16 \times 9 = 144 \) and add to \( -24 \). These numbers are \( -12 \) and \( -12 \):
\[ 16a^2 - 12ab - 12ab + 9b^2 \]
Now, we can group the terms:
\[ (16a^2 - 12ab) + (-12ab + 9b^2) \]
Factoring by grouping gives:
\[ 4a(4a - 3b) - 3b(4a - 3b) \]
This can be factored further:
\[ (4a - 3b)(4a - 3b) = (4a - 3b)^2 \]

### Step 6: Substitute back \( a \) and \( b \)
Now, substituting back \( a = 2x - y \) and \( b = 2x + y \):
\[ (4(2x - y) - 3(2x + y))^2 \]
Simplifying this gives:
\[ (8x - 4y - 6x - 3y)^2 = (2x - 7y)^2 \]

### Final Answer
Thus, the factorized form of the original expression is:
\[ (2x - 7y)^2 \]