jmc

By Vieta's formulas, $$\begin{array}{rl}a+b+c& =6,\\ ab+ac+bc& =3,\\ abc& =-1.\end{array}$$Let $p=a^{2}b+b^{2}c+c^{2}a$ and $q=a{b}^2+b{c}^2+c{a}^2.$ Then $$p+q=a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2.$$Note that $$(a+b+c)(ab+ac+bc)=a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2+3abc,$$so $$\begin{array}{rl}{a}^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2& =(a+b+c)(ab+ac+bc)-3abc\\ & =(6)(3)-3(-1)\\ & =21.\end{array}$$Also, $$pq=a^3{b}^3+a^3{c}^3+b^3{c}^3+a^{4}bc+a{b}^{4}c+ab{c}^4+3a^2{b}^2{c}^2.$$To obtain the terms $a^3{b}^3+a^3{c}^3+b^3{c}^3,$ we can cube $ab+ac+bc$: $$\begin{array}{rl}(ab+ac+bc{)}^3& =a^3{b}^3+a^3{c}^3+b^3{c}^3\\ & +3(a^3{b}^{2}c+a^{3}b{c}^2+a^2{b}^{3}c+a^{2}b{c}^3+a{b}^3{c}^2+a{b}^2{c}^3)\\ & +6a^2{b}^2{c}^2.\end{array}$$Now, $$\begin{array}{rl}& a^3{b}^{2}c+a^{3}b{c}^2+a^2{b}^{3}c+a^{2}b{c}^3+a{b}^3{c}^2+a{b}^2{c}^3\\ & =abc(a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2)\\ & =(-1)(21)=-21,\end{array}$$so $$\begin{array}{rl}{a}^3{b}^3+a^3{c}^3+b^3{c}^3& =(ab+ac+bc{)}^3-3(-21)-6a^2{b}^2{c}^2\\ & =3^3-3(-21)-6(-1{)}^2\\ & =84.\end{array}$$Also, $$a^{4}bc+a{b}^{4}c+ab{c}^4=abc(a^3+b^3+c^3).$$To obtain the terms $a^3+b^3+c^3,$ we can cube $a+b+c$: $$(a+b+c{)}^3=a^3+b^3+c^3+3(a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2)+6abc,$$so $$\begin{array}{rl}{a}^3+b^3+c^3& =(a+b+c{)}^3-3(a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2)-6abc\\ & =6^3-3(21)-6(-1)\\ & =159.\end{array}$$Hence, $$\begin{array}{rl}pq& =a^3{b}^3+a^3{c}^3+b^3{c}^3+a^{4}bc+a{b}^{4}c+ab{c}^4+3a^2{b}^2{c}^2\\ & =84+(-1)(159)+3(-1{)}^2\\ & =-72.\end{array}$$Then by Vieta's formulas, $p$ and $q$ are the roots of $$x^2-21x-72=(x-24)(x+3)=0.$$Thus, the possible values of $p$ (and $q$) are $\boxed{24,-3}.$