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

Answer

To factorise the expression \( (x^2 + 4y^2 - 9z^2)^2 - 16x^2y^2 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (x^2 + 4y^2 - 9z^2)^2 - 16x^2y^2 \]
Notice that \( 16x^2y^2 \) can be rewritten as \( (4xy)^2 \). Thus, we have:
\[ (x^2 + 4y^2 - 9z^2)^2 - (4xy)^2 \]

### Step 2: Apply the difference of squares formula
This expression is in the form of \( a^2 - b^2 \), where:
- \( a = x^2 + 4y^2 - 9z^2 \)
- \( b = 4xy \)

Using the difference of squares identity \( a^2 - b^2 = (a + b)(a - b) \), we can factor it as follows:
\[ (x^2 + 4y^2 - 9z^2 + 4xy)(x^2 + 4y^2 - 9z^2 - 4xy) \]

### Step 3: Simplify each factor
Now we simplify each factor:
1. The first factor:
\[ x^2 + 4y^2 - 9z^2 + 4xy \]
This can be rearranged as:
\[ x^2 + 4xy + 4y^2 - 9z^2 = (x + 2y)^2 - 9z^2 \]

2. The second factor:
\[ x^2 + 4y^2 - 9z^2 - 4xy \]
This can be rearranged as:
\[ x^2 - 4xy + 4y^2 - 9z^2 = (x - 2y)^2 - 9z^2 \]

### Step 4: Apply the difference of squares again
Now we have two factors:
1. \( (x + 2y)^2 - (3z)^2 \)
2. \( (x - 2y)^2 - (3z)^2 \)

Again, we can apply the difference of squares formula:
1. For \( (x + 2y)^2 - (3z)^2 \):
\[ (x + 2y - 3z)(x + 2y + 3z) \]

2. For \( (x - 2y)^2 - (3z)^2 \):
\[ (x - 2y - 3z)(x - 2y + 3z) \]

### Final Factorization
Combining all these factors, we get the final factorization of the original expression:
\[ (x + 2y - 3z)(x + 2y + 3z)(x - 2y - 3z)(x - 2y + 3z) \]