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 $24,-3.$
베트남의 공식으로,
\begin{align*}
a + b + c &= 6, \\.
ab + ac + bc &= 3, \.
abc &= -1.
\end{align*}$p = a^2 b + b^2 c + c^2 a$ 및 $q = ab^2 + bc^2 + ca^2$라고 하자.
\$[p + q = a^2 b + ab^2 + a^2 c + ac^2 + b^2 c + bc^2.$]에 유의하십시오.
\[(a + b + c)(ab + ac + bc) = a^2 b + ab^2 + a^2 c + ac^2 + b^2 c + bc^2 + 3abc,\]따라서 다음과 같습니다.
\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*}또한,
\[pq = a^3 b^3 + a^3 c^3 + b^3 c^3 + a^4 bc + ab^4 c + abc^4 + 3a^2 b^2 c^2.\]항 $a^3 b^3 + a^3 c^3 + b^3 c^3,$를 구하려면 $ab + ac + bc$를 세제곱하면 됩니다:
\$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*}지금,
\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*}또한,
\[a^4 bc + ab^4 c + abc^4 = abc(a^3 + b^3 + c^3).\]항 $a^3 + b^3 + c^3,$를 구하려면 $a + b + c$를 세제곱하면 됩니다:
\$(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,$]그래서 다음과 같습니다.
\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*}따라서,
\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*}그렇다면 Vieta의 공식에 따라 $p$와 $q$는 다음의 근이 됩니다.
\[x^2 - 21x - 72 = (x - 24)(x + 3) = 0.\]따라서 $p$(및 $q$)의 가능한 값은 $24,-3.$입니다.