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

### Step 1: Expand the squares
We start by expanding \((x + y + z)^{2}\) and \((x - y - z)^{2}\).

\[ (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2xz + 2yz \]

\[ (x - y - z)^{2} = x^{2} + y^{2} + z^{2} - 2xy - 2xz - 2yz \]

### Step 2: Substitute the expansions into the expression
Now we substitute these expansions into the original expression:

\[ (x + y + z)^{2} - (x - y - z)^{2} + 4y^{2} - 4z^{2} \]

This becomes:

\[ \left( x^{2} + y^{2} + z^{2} + 2xy + 2xz + 2yz \right) - \left( x^{2} + y^{2} + z^{2} - 2xy - 2xz - 2yz \right) + 4y^{2} - 4z^{2} \]

### Step 3: Simplify the expression
Now, we simplify the expression:

\[ = (x^{2} + y^{2} + z^{2} + 2xy + 2xz + 2yz) - (x^{2} + y^{2} + z^{2}) + 2xy + 2xz + 2yz + 4y^{2} - 4z^{2} \]

The \(x^{2}\), \(y^{2}\), and \(z^{2}\) terms cancel out:

\[ = 2xy + 2xz + 2yz + 4y^{2} - 4z^{2} \]

### Step 4: Combine like terms
Now we can combine the terms:

\[ = 2xy + 2xz + 2yz + 4(y^{2} - z^{2}) \]

### Step 5: Factor out common terms
We can factor out a \(2\) from the first three terms and recognize that \(y^{2} - z^{2}\) is a difference of squares:

\[ = 2(xy + xz + 2(y^{2} - z^{2})) \]

\[ = 2(xy + xz + 4(y + z)(y - z)) \]

### Step 6: Final factorization
Now we can factor \(y^{2} - z^{2}\) as \((y + z)(y - z)\):

\[ = 2(xy + xz + 4(y + z)(y - z)) \]

Thus, the complete factorization of the expression is:

\[ = 4(y + z)(x + y - z) \]

### Final Answer:
\[ = 4(y + z)(x + y - z) \]