### Instruction: Factor
\[\frac{(a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3}{(a - b)^3 + (b - c)^3 + (c - a)^3}.\].### Response: I recognize this expression as a special case of a well-known identity that says that if $x + y + z = 0$, then $x^3 + y^3 + z^3 = 3xyz$.
This identity can be proved by expanding $(x + y + z)^3$ and simplifying.
In this problem, I can let $x = a^2 - b^2, y = b^2 - c^2, z = c^2 - a^2$, and I can see that $x + y + z = 0$.
Therefore, by the identity, I have that $(a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3 = 3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2)$.
Similarly, I can let $x = a - b, y = b - c, z = c - a$, and I can see that $x + y + z = 0$.
Therefore, by the identity, I have that $(a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a)$.
Now, I can divide the numerator and the denominator by 3, and I get that the original expression is equal to $\frac{(a^2 - b^2)(b^2 - c^2)(c^2 - a^2)}{(a - b)(b - c)(c - a)}$.
This looks like a nice factorization, but I can still simplify it further by noticing that $(a^2 - b^2) = (a - b)(a + b)$, and similarly for the other terms.
So, I can rewrite the expression as $\frac{(a - b)(a + b)(b - c)(b + c)(c - a)(c + a)}{(a - b)(b - c)(c - a)}$.
Now, I can cancel out the common factors of $(a - b), (b - c), (c - a)$ in the numerator and the denominator, and I get that the expression is equal to $(a + b)(b + c)(c + a)$.
This is the simplest form of the expression, and I have factored it completely.