jmc

First, we take out a factor of $a-b$: $$\begin{array}{rl}{a}^3(b^2-c^2)+b^3(c^2-a^2)+c^3(a^2-b^2)& =a^3{b}^2-a^2{b}^3+b^3{c}^2-a^3{c}^2+c^3(a+b)(a-b)\\ & =a^2{b}^2(a-b)+(b^3-a^3)c^2+c^3(a+b)(a-b)\\ & =(a-b)[a^2{b}^2-(a^2+ab+b^2)c^2+c^3(a+b)]\\ & =(a-b)(a^2{b}^2-a^2{c}^2-ab{c}^2-b^2{c}^2+a{c}^3+b{c}^3).\end{array}$$We can then take out a factor of $b-c$: $$\begin{array}{rl}{a}^2{b}^2-a^2{c}^2-ab{c}^2-b^2{c}^2+a{c}^3+b{c}^3& =a^2(b^2-c^2)+a{c}^3-ab{c}^2+b{c}^3-b^2{c}^2\\ & =a^2(b^2-c^2)+a{c}^2(c-b)+b{c}^2(c-b)\\ & =a^2(b-c)(b+c)+a{c}^2(c-b)+b{c}^2(c-b)\\ & =(b-c)[a^2(b+c)-a{c}^2-b{c}^2]\\ & =(b-c)(a^{2}b+a^{2}c-a{c}^2-b{c}^2).\end{array}$$Finally, we take out a factor of $c-a$: $$\begin{array}{rl}{a}^{2}b+a^{2}c-a{c}^2-b{c}^2& =a^{2}b-b{c}^2+a^{2}c-a{c}^2\\ & =b(a^2-c^2)+ac(a-c)\\ & =b(a-c)(a+c)+ac(a-c)\\ & =-(c-a)(ab+ac+bc).\end{array}$$Thus, $p(a,b,c)=\boxed{-(ab+ac+bc)}.$