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

Answer

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

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

### Step 2: Recognize the difference of squares
The expression now resembles the difference of squares, which is given by the formula:
\[ a^2 - b^2 = (a - b)(a + b) \]
In our case, let:
- \( a = x^2 + y^2 - z^2 \)
- \( b = 2xy \)

### Step 3: Apply the difference of squares formula
Now we can apply the difference of squares formula:
\[ (x^2 + y^2 - z^2 - 2xy)(x^2 + y^2 - z^2 + 2xy) \]

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

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

### Final Result
Putting it all together, we have:
\[ (x^2 + y^2 - z^2)^2 - 4x^2y^2 = ((x - y) - z)((x - y) + z)((x + y) - z)((x + y) + z) \]