Simplify: `((x^(2)-y^(2))^(3) + (y^(2) - z^(2))^(3) + (z^(2) -x^(2))^(3))/((x-y)^(3) + (y-z)^(3) + (z-x)^(3))`

Answer

To simplify 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 follow these steps:

### Step 1: Identify the terms in the numerator and denominator
The numerator consists of three cubes:
- \( (x^2 - y^2)^3 \)
- \( (y^2 - z^2)^3 \)
- \( (z^2 - x^2)^3 \)

The denominator also consists of three cubes:
- \( (x - y)^3 \)
- \( (y - z)^3 \)
- \( (z - x)^3 \)

### Step 2: Use the identity for sums of cubes
We can apply the identity for the sum of cubes, which states that if \( a + b + c = 0 \), then

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

### Step 3: Check if the sum of the terms in the numerator equals zero
Let:
- \( a = x^2 - y^2 \)
- \( b = y^2 - z^2 \)
- \( c = z^2 - x^2 \)

Calculating \( a + b + c \):

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

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

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

### Step 4: Check if the sum of the terms in the denominator equals zero
Let:
- \( p = x - y \)
- \( q = y - z \)
- \( r = z - x \)

Calculating \( p + q + r \):

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

Since \( p + q + r = 0 \), we can use the identity:

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

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

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

### Step 6: Cancel the common factor
The 3s in the numerator and denominator cancel out:

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

### Step 7: Factor the differences of squares
Using the difference of squares, we can rewrite \( x^2 - y^2 \) as \( (x - y)(x + y) \), and similarly for the other terms:

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

### Step 8: Cancel the common factors
The common factors \( (x - y) \), \( (y - z) \), and \( (z - x) \) in the numerator and denominator cancel out:

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

### Final Answer
Thus, the simplified expression is:

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

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