Find `(x+y+z)^(3)-(x+y-z)^(3)-(y+z-x)^(3) -(z + x-y)^(3)`

Options

A. 24xyz
B. 27 xyz
C. 3xyz
D. 0

Answer

To solve the expression \((x+y+z)^{3} - (x+y-z)^{3} - (y+z-x)^{3} - (z+x-y)^{3}\), we can use the properties of cubes and some algebraic manipulation.

### Step-by-step Solution:

1. **Identify the Expression**:
We start with the expression:
\[ E = (x+y+z)^{3} - (x+y-z)^{3} - (y+z-x)^{3} - (z+x-y)^{3} \]

2. **Use the Identity for Cubes**:
Recall the identity for the difference of cubes:
\[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]
We can apply this identity to each of the terms in our expression.

3. **Calculate Each Term**:
- Let \(a = x+y+z\) and \(b = x+y-z\):
\[ E_1 = (x+y+z)^{3} - (x+y-z)^{3} = (2z)((x+y+z)^{2} + (x+y+z)(x+y-z) + (x+y-z)^{2}) \]
- Let \(a = y+z-x\) and \(b = z+x-y\):
\[ E_2 = (y+z-x)^{3} - (z+x-y)^{3} = (2x)((y+z-x)^{2} + (y+z-x)(z+x-y) + (z+x-y)^{2}) \]

4. **Combine the Results**:
Now, we can combine the results from \(E_1\) and \(E_2\):
\[ E = 2z \cdot ((x+y+z)^{2} + (x+y+z)(x+y-z) + (x+y-z)^{2}) - 2x \cdot ((y+z-x)^{2} + (y+z-x)(z+x-y) + (z+x-y)^{2}) \]

5. **Evaluate at Specific Values**:
To simplify further, we can substitute specific values for \(x\), \(y\), and \(z\). For instance, let \(x = 1\), \(y = 2\), and \(z = 3\):
\[ E = (1+2+3)^{3} - (1+2-3)^{3} - (2+3-1)^{3} - (3+1-2)^{3} \]
\[ = 6^{3} - 0^{3} - 4^{3} - 2^{3} \]
\[ = 216 - 0 - 64 - 8 \]
\[ = 216 - 72 = 144 \]

### Final Result:
Thus, the value of the expression is:
\[ E = 144 \]