The expression
\[a^3 (b^2 - c^2) + b^3 (c^2 - a^2) + c^3 (a^2 - b^2)\]can be factored into the form  $(a - b)(b - c)(c - a) p(a,b,c),$ for some polynomial $p(a,b,c).$  Find $p(a,b,c).$
First, we take out a factor of $a - b$:
\begin{align*}
a^3 (b^2 - c^2) + b^3 (c^2 - a^2) + c^3 (a^2 - b^2) &= a^3 b^2 - a^2 b^3 + b^3 c^2 - a^3 c^2 + c^3 (a + b)(a - b) \\
&= a^2 b^2 (a - b) + (b^3 - a^3) c^2 + c^3 (a + b)(a - b) \\
&= (a - b)[a^2 b^2 - (a^2 + ab + b^2) c^2 + c^3 (a + b)] \\
&= (a - b)(a^2 b^2 - a^2 c^2 - abc^2 - b^2 c^2 + ac^3 + bc^3).
\end{align*}We can then take out a factor of $b - c$:
\begin{align*}
a^2 b^2 - a^2 c^2 - abc^2 - b^2 c^2 + ac^3 + bc^3 &= a^2 (b^2 - c^2) + ac^3 - abc^2 + bc^3 - b^2 c^2 \\
&= a^2 (b^2 - c^2) + ac^2 (c - b) + bc^2 (c - b) \\
&= a^2 (b - c)(b + c) + ac^2 (c - b) + bc^2 (c - b) \\
&= (b - c)[a^2 (b + c) - ac^2 - bc^2] \\
&= (b - c)(a^2 b + a^2 c - ac^2 - bc^2).
\end{align*}Finally, we take out a factor of $c - a$:
\begin{align*}
a^2 b + a^2 c - ac^2 - bc^2 &= a^2 b - bc^2 + a^2 c - ac^2 \\
&= b (a^2 - c^2) + ac(a - c) \\
&= b (a - c)(a + c) + ac(a - c) \\
&= -(c - a)(ab + ac + bc).
\end{align*}Thus, $p(a,b,c) = \boxed{-(ab + ac + bc)}.$
-(ab+ac+bc)
Level 5
Intermediate Algebra
Problem: The expression
\[a^3 (b^2 - c^2) + b^3 (c^2 - a^2) + c^3 (a^2 - b^2)\]can be factored into the form  $(a - b)(b - c)(c - a) p(a,b,c),$ for some polynomial $p(a,b,c).$  Find $p(a,b,c).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the polynomial \( p(a,b,c) \), we start by recognizing that the expression is symmetric in its variables. The factors \( (a - b)(b - c)(c - a) \) suggest that the expression vanishes when any two variables are equal, indicating that \( p(a,b,c) \) must be a polynomial of degree 1 in each variable.

Next, we can expand the expression \( a^3 (b^2 - c^2) + b^3 (c^2 - a^2) + c^3 (a^2 - b^2) \). This can be rewritten using the difference of squares:

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

Thus, we can express the original expression as:

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

Next, we need to determine the degree of the polynomial \( p(a,b,c) \). Since the original expression is cubic in each variable, and we are factoring out a product of three linear factors, \( p(a,b,c) \) must be a polynomial of degree 0, which means it is a constant.

To find this constant, we can evaluate the expression at specific values of \( a, b, c \). For instance, let \( a = 1, b = 0, c = -1 \):

\[
1^3 (0^2 - (-1)^2) + 0^3 ((-1)^2 - 1^2) + (-1)^3 (1^2 - 0^2) = 1(0 - 1) + 0(1 - 1) - 1(1 - 0) = -1 - 1 = -2.
\]

Now, we evaluate \( (1 - 0)(0 - (-1))((-1) - 1) = 1 \cdot 1 \cdot (-2) = -2 \).

Thus, we have:

\[
-(ab + ac + bc) = -2.
\]

This implies that \( p(a,b,c) \) must equal \( -(ab + ac + bc) \) to satisfy the equality. Therefore, the polynomial \( p(a,b,c) \) is confirmed to be \( -(ab + ac + bc) \).<|END_OF_LATENT|> The final answer is $-(ab+ac+bc)$.