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 use the identity for the sum of cubes and some algebraic manipulation.

### Step-by-Step Solution:

**Step 1: Identify the Numerator and Denominator**

We have:

Numerator: \((a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3\)

Denominator: \((a - b)^3 + (b - c)^3 + (c - a)^3\)

**Step 2: Use the Sum of Cubes Identity**

Recall the identity:

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

If \(x + y + z = 0\), then:

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

**Step 3: Set \(x = a^2 - b^2\), \(y = b^2 - c^2\), \(z = c^2 - a^2\)**

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

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

This simplifies to:

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

So, we can apply the identity:

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

**Step 4: Apply the Identity to the Denominator**

Now, let \(x = a - b\), \(y = b - c\), \(z = c - a\). We check if \(x + y + z = 0\):

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

Thus, we can apply the same identity:

\[ (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a) \]

**Step 5: Substitute Back into the Expression**

Now substituting back into our original expression:

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

The \(3\) cancels out:

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

**Step 6: Factor the Numerator**

We can factor \(a^2 - b^2\) as \((a - b)(a + b)\), \(b^2 - c^2\) as \((b - c)(b + c)\), and \(c^2 - a^2\) as \((c - a)(c + a)\):

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

**Step 7: Cancel Common Factors**

Now, we can cancel the common factors \((a - b)\), \((b - c)\), and \((c - a)\):

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

### Final Answer:

Thus, the simplified expression is:

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