By Vieta's formulas, \begin{align*} a + b + c &= 6, \\ ab + ac + bc &= 3, \\ abc &= -1.\end{align*}Let $ p = a^2 b + b^2 c + c^2 a $ and $ q = ab^2 + bc^2 + ca^2 $. Then \[p + q = a^2 b + ab^2 + a^2 c + ac^2 + b^2 c + bc^2.\]Note that \[(a + b + c)(ab + ac + bc) = a^2 b + ab^2 + a^2 c + ac^2 + b^2 c + bc^2 + 3abc,\]so \begin{align*} a^2 b + ab^2 + a^2 c + ac^2 + b^2 c + bc^2 &= (a + b + c)(ab + ac + bc) - 3abc \\ &= (6)(3) - 3(-1) \\ &= 21.\end{align*}Also, \[pq = a^3 b^3 + a^3 c^3 + b^3 c^3 + a^4 bc + ab^4 c + abc^4 + 3a^2 b^2 c^2.\]To obtain the terms $ a^3 b^3 + a^3 c^3 + b^3 c^3, $ we can cube $ ab + ac + bc $: \begin{align*} (ab + ac + bc)^3 &= a^3 b^3 + a^3 c^3 + b^3 c^3 \\ &\quad + 3(a^3 b^2 c + a^3 bc^2 + a^2 b^3 c + a^2 bc^3 + ab^3 c^2 + ab^2 c^3) \\ &\quad + 6a^2 b^2 c^2.\end{align*}Now, \begin{align*} &a^3 b^2 c + a^3 bc^2 + a^2 b^3 c + a^2 bc^3 + ab^3 c^2 + ab^2 c^3 \\ &= abc (a^2 b + ab^2 + a^2 c + ac^2 + b^2 c + bc^2) \\ &= (-1)(21) = -21, \end{align*}so \begin{align*} 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{align*}Also, \[a^4 bc + ab^4 c + abc^4 = abc(a^3 + b^3 + c^3).\]To obtain the terms $ a^3 + b^3 + c^3, $ we can cube $ a + b + c $: \[(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a^2 b + ab^2 + a^2 c + ac^2 + b^2 c + bc^2) + 6abc,\]so \begin{align*} a^3 + b^3 + c^3 &= (a + b + c)^3 - 3(a^2 b + ab^2 + a^2 c + ac^2 + b^2 c + bc^2) - 6abc \\ &= 6^3 - 3(21) - 6(-1) \\ &= 159.\end{align*}Hence, \begin{align*} pq &= a^3 b^3 + a^3 c^3 + b^3 c^3 + a^4 bc + ab^4 c + abc^4 + 3a^2 b^2 c^2 \\ &= 84 + (-1)(159) + 3(-1)^2 \\ &= -72.\end{align*}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} $.