Simplify `((a^(2)-b^(2))^(3)+(b^(2)-c^(2))^(3)+(c^(2)-a^(2))^(3))/((a-b)^(3)+(b-c)^(3)+(c-a)^(3))`

Answer

To simplify the expression

\[ \frac{(a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3}{(a - b)^3 + (b - c)^3 + (c - a)^3} \]

we can follow these steps:

### Step 1: Identify the Form of the Numerator and Denominator
Both the numerator and denominator are in the form of a sum of cubes. We can use the identity:

\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) \]

if \(x + y + z = 0\).

### Step 2: Set Up the Variables
Let:
- \(x = a^2 - b^2\)
- \(y = b^2 - c^2\)
- \(z = c^2 - a^2\)

Now, we check if \(x + y + z = 0\):

\[ (a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) = a^2 - b^2 + b^2 - c^2 + c^2 - a^2 = 0 \]

Since \(x + y + z = 0\), we can apply the identity.

### Step 3: Apply the Identity to the Numerator
Using the identity, we have:

\[ x^3 + y^3 + z^3 = 3xyz \]

Thus, the numerator becomes:

\[ 3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2) \]

### Step 4: Set Up the Denominator
Now, let’s apply the same logic to the denominator. Let:

- \(u = a - b\)
- \(v = b - c\)
- \(w = c - a\)

Again, we check if \(u + v + w = 0\):

\[ (a - b) + (b - c) + (c - a) = a - b + b - c + c - a = 0 \]

So we can apply the identity here as well:

\[ u^3 + v^3 + w^3 = 3uvw \]

Thus, the denominator becomes:

\[ 3(a - b)(b - c)(c - a) \]

### Step 5: Substitute Back into the Expression
Now we substitute the simplified forms of the numerator and denominator back into the original expression:

\[ \frac{3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2)}{3(a - b)(b - c)(c - a)} \]

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

\[ \frac{(a^2 - b^2)(b^2 - c^2)(c^2 - a^2)}{(a - b)(b - c)(c - a)} \]

### Step 7: Factor the Numerator
We can factor \(a^2 - b^2\) using the difference of squares:

\[ a^2 - b^2 = (a - b)(a + b) \]
\[ b^2 - c^2 = (b - c)(b + c) \]
\[ c^2 - a^2 = (c - a)(c + a) \]

Substituting these back into the expression gives:

\[ \frac{(a - b)(a + b)(b - c)(b + c)(c - a)(c + a)}{(a - b)(b - c)(c - a)} \]

### Step 8: Cancel Common Terms
Now we can cancel \((a - b)\), \((b - c)\), and \((c - a)\):

\[ (a + b)(b + c)(c + a) \]

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

\[ (a + b)(b + c)(c + a) \]