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

Options

A. `(x+ 2z) (4x + y -z)`
B. `(x+z) (x+y-2z)`
C. `4(y+z) (x+y - z)`
D. `2(x+z) (y-2z)`

Answer

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

### Step 1: Identify the structure
We recognize that the expression has the form of a difference of squares, which is given by the formula \(a^2 - b^2 = (a + b)(a - b)\).

### Step 2: Apply the difference of squares
Let:
- \(a = (x + y + z)\)
- \(b = (x - y - z)\)

Now, we can rewrite the first part of the expression:
\[ (x + y + z)^2 - (x - y - z)^2 = (a + b)(a - b) \]

### Step 3: Calculate \(a + b\) and \(a - b\)
Calculating \(a + b\):
\[ a + b = (x + y + z) + (x - y - z) = 2x + 0 + 0 = 2x \]

Calculating \(a - b\):
\[ a - b = (x + y + z) - (x - y - z) = 0 + 2y + 2z = 2y + 2z \]

### Step 4: Substitute back into the expression
Now substituting back, we have:
\[ (x + y + z)^2 - (x - y - z)^2 = (2x)(2y + 2z) = 4x(y + z) \]

### Step 5: Factor the remaining part
Now, we need to factor the remaining part \(4y^2 - 4z^2\):
This can also be factored using the difference of squares:
\[ 4y^2 - 4z^2 = 4(y^2 - z^2) = 4(y - z)(y + z) \]

### Step 6: Combine the factors
Now we combine both parts:
\[ 4x(y + z) + 4(y - z)(y + z) \]

### Step 7: Factor out the common term
Notice that \( (y + z) \) is a common factor:
\[ = 4(y + z)(x + (y - z)) \]

### Final Answer
Thus, the factorised form of the expression is:
\[ \boxed{4(y + z)(x + y - z)} \]