### Instruction: The expression
\[a(b - c)^3 + b(c - a)^3 + c(a - b)^3\]can be factored into the form  $(a - b)(b - c)(c - a) p(a,b,c),$ for some polynomial $p(a,b,c).$  Find $p(a,b,c).$.### Response: I notice that the expression has a symmetric structure: if I swap any two variables, the expression does not change.
This suggests that the polynomial $p(a,b,c)$ might also be symmetric, or at least have some symmetry properties.
To test this idea, I will try to find the value of $p(a,b,c)$ for some special cases where two or more variables are equal.
If I set $a = b,$ then the expression becomes $a(b - c)^3 + a(c - a)^3 + c(a - a)^3 = 0,$ by simplifying the terms.
This means that $(a - b)$ is a factor of the expression, and also that $p(a,a,c) = 0.$
Similarly, if I set $b = c,$ then the expression becomes $a(b - c)^3 + c(c - a)^3 + c(a - b)^3 = 0,$ and $p(a,b,b) = 0.$
And if I set $c = a,$ then the expression becomes $a(b - a)^3 + b(a - a)^3 + a(a - b)^3 = 0,$ and $p(a,b,a) = 0.$
So, I have found that $p(a,b,c)$ is zero whenever two variables are equal.
This implies that $p(a,b,c)$ is divisible by $(a - b)(b - c)(c - a),$ which are the same factors as in the original expression.
Therefore, $p(a,b,c)$ must be a constant, say $k.$
To find the value of $k,$ I will plug in some arbitrary values for $a, b, c$ and see what the expression becomes.
For example, if I let $a = 1, b = 2, c = 3,$ then the expression becomes $1(2 - 3)^3 + 2(3 - 1)^3 + 3(1 - 2)^3 = -1 + 16 - 9 = 6.$
This means that $6 = (1 - 2)(2 - 3)(3 - 1) k,$ which simplifies to $6 = 2k.$
So, $k = 3,$ and $p(a,b,c) = 3.$