jmc

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