`(x-y-z)^(2)-(x+y+z)^(2)` is equal to

Options

A. `4xy+4yz`
B. `-4xy-4xz`
C. `4xy+4xz`
D. `-4yz`

Answer

To solve the expression \((x - y - z)^2 - (x + y + z)^2\), we can use the algebraic identities for the square of a binomial. Let's break it down step by step.

### Step 1: Expand both squares using the identity
The identity for the square of a binomial states that \((a - b)^2 = a^2 - 2ab + b^2\) and \((a + b)^2 = a^2 + 2ab + b^2\).

1. **Expand \((x - y - z)^2\)**:
\[ (x - y - z)^2 = x^2 - 2x(y + z) + (y + z)^2 \]
Expanding \((y + z)^2\):
\[ (y + z)^2 = y^2 + 2yz + z^2 \]
Thus,
\[ (x - y - z)^2 = x^2 - 2xy - 2xz + y^2 + 2yz + z^2 \]

2. **Expand \((x + y + z)^2\)**:
\[ (x + y + z)^2 = x^2 + 2x(y + z) + (y + z)^2 \]
Again, expanding \((y + z)^2\):
\[ (y + z)^2 = y^2 + 2yz + z^2 \]
Thus,
\[ (x + y + z)^2 = x^2 + 2xy + 2xz + y^2 + 2yz + z^2 \]

### Step 2: Subtract the two expansions
Now we will subtract the second expansion from the first:
\[ (x - y - z)^2 - (x + y + z)^2 = (x^2 - 2xy - 2xz + y^2 + 2yz + z^2) - (x^2 + 2xy + 2xz + y^2 + 2yz + z^2) \]

### Step 3: Simplify the expression
Now, we will simplify the expression:
- The \(x^2\), \(y^2\), \(z^2\), and \(2yz\) terms cancel out:
\[ = -2xy - 2xz - 2xy - 2xz \]
Combining like terms:
\[ = -4xy - 4xz \]

### Final Result
Thus, the expression \((x - y - z)^2 - (x + y + z)^2\) simplifies to:
\[ \boxed{-4xy - 4xz} \]