The value of the expression
`((x ^(2) - y ^(2)) ^(3) + ( y ^(2) - z ^(2)) ^(3) + (z ^(2) - x ^(2)) ^(3))/((x - y) ^(3) + ( y - z) ^(3) + (z - x ) ^(3))` is

Options

A. `(x ^(2) - y ^(2)) (y ^(2) - z ^(2)) ( z ^(2) - x ^(2))`
B. `3 (x - x ) ( y -z) (z -x)`
C. `(x + y) (y + z) (z + x)`
D. `(3 (x + y) (y + z) (z + x)`

Answer

To find the value of the expression

\[ \frac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y - z)^3 + (z - x)^3}, \]

we can use the identity for the sum of cubes.

### Step 1: Identify the terms
Let:
- \( a = x^2 - y^2 \)
- \( b = y^2 - z^2 \)
- \( c = z^2 - x^2 \)

### Step 2: Check if \( a + b + c = 0 \)
Now, we calculate \( a + b + c \):

\[ a + b + c = (x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2) = x^2 - y^2 + y^2 - z^2 + z^2 - x^2 = 0. \]

Since \( a + b + c = 0 \), we can use the identity for the sum of cubes:

\[ a^3 + b^3 + c^3 = 3abc. \]

### Step 3: Apply the identity to the numerator
Thus, we have:

\[ (x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3 = 3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2). \]

### Step 4: Apply the identity to the denominator
Now, let’s consider the denominator:

Let:
- \( p = x - y \)
- \( q = y - z \)
- \( r = z - x \)

Again, we check if \( p + q + r = 0 \):

\[ p + q + r = (x - y) + (y - z) + (z - x) = 0. \]

Using the same identity for the sum of cubes:

\[ (x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x). \]

### Step 5: Substitute back into the expression
Now substituting back into our original expression, we get:

\[ \frac{3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)}{3(x - y)(y - z)(z - x)}. \]

### Step 6: Simplify the expression
The \( 3 \) cancels out:

\[ \frac{(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)}{(x - y)(y - z)(z - x)}. \]

### Step 7: Factor the numerator
Notice that \( x^2 - y^2 = (x - y)(x + y) \), \( y^2 - z^2 = (y - z)(y + z) \), and \( z^2 - x^2 = (z - x)(z + x) \).

Thus, we can rewrite the numerator:

\[ (x - y)(x + y)(y - z)(y + z)(z - x)(z + x). \]

### Step 8: Cancel common factors
Now, substituting this back into our expression, we have:

\[ \frac{(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)}{(x - y)(y - z)(z - x)}. \]

The common factors \( (x - y) \), \( (y - z) \), and \( (z - x) \) cancel out:

\[ (x + y)(y + z)(z + x). \]

### Final Answer
Thus, the value of the expression is:

\[ (x + y)(y + z)(z + x). \]