Given that $a$, $b$, and $c$ are three distinct real numbers. If the quadratic equations
$$
\begin{array}{l}
x^{2} + a x + b = 0, \\
x^{2} + b x + c = 0, \\
x^{2} + c x + a = 0
\end{array}
$$

each have exactly one common root with any other equation, find the value of $a^{2} + b^{2} + c^{2}$. ${ }^{[1]}$
Let the equations (1) and (3) have only one common root $x_{1}$, equations (1) and (2) have only one common root $x_{2}$, and equations (2) and (3) have only one common root $x_{3}$. Therefore, the roots of equation (1) are $x_{1}$ and $x_{2}$, the roots of equation (2) are $x_{2}$ and $x_{3}$, and the roots of equation (3) are $x_{3}$ and $x_{1}$.

Since $x_{1}$ is a root of equations (1) and (3), we have
$$
\left\{\begin{array}{l}
x_{1}^{2} + a x_{1} + b = 0, \\
x_{1}^{2} + c x_{1} + a = 0
\end{array} \Rightarrow x_{1} = \frac{a - b}{a - c} .\right.
$$

Similarly, $x_{2} = \frac{b - c}{b - a}$ and $x_{3} = \frac{c - a}{c - b}$. Thus, $x_{1} x_{2} x_{3} = -1$.
Clearly, $x_{1}$, $x_{2}$, and $x_{3}$ are distinct. Otherwise, suppose $x_{2} = x_{3}$.
$$
\begin{array}{l}
\text { Hence } \frac{b - c}{b - a} = \frac{c - a}{c - b} \\
\Rightarrow a^{2} + b^{2} + c^{2} - b c - c a - a b = 0 \\
\Rightarrow (b - c)^{2} + (c - a)^{2} + (a - b)^{2} = 0 \\
\Rightarrow a = b = c,
\end{array}
$$

which contradicts the given conditions.
Thus, $x_{1}$, $x_{2}$, and $x_{3}$ are distinct.
By Vieta's formulas, we have
$$
\begin{array}{l}
a = -x_{1} - x_{2} = x_{3} x_{1}, \\
b = -x_{2} - x_{3} = x_{1} x_{2}, \\
c = -x_{3} - x_{1} = x_{2} x_{3}.
\end{array}
$$

From the above three equations, we get
$$
\begin{array}{l}
-2\left(x_{1} + x_{2} + x_{3}\right) = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2}, \text { (4) } \\
x_{1}\left(1 + x_{3}\right) x_{2}\left(1 + x_{1}\right) x_{3}\left(1 + x_{2}\right) \\
= \left(-x_{2}\right)\left(-x_{3}\right)\left(-x_{1}\right).
\end{array}
$$

Noting that $x_{1} x_{2} x_{3} = -1$,
we have $\left(1 + x_{3}\right)\left(1 + x_{1}\right)\left(1 + x_{2}\right) = -1$
$\Rightarrow x_{1} + x_{2} + x_{3} + x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -1$.
From equations (4) and (5), we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1} + x_{2} + x_{3} = 1, \\
x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a + b + c = -2, \\
b c + c a + a b = -1, \\
a b c = 1
\end{array}\right. \\
\Rightarrow a^{2} + b^{2} + c^{2} \\
= (a + b + c)^{2} - 2(b c + c a + a b) \\
\quad = (-2)^{2} - 2(-1) = 6.
\end{array}
$$

Let's explore further.
From the above solution, we also obtained
$$
\left\{\begin{array}{l}
x_{1} + x_{2} + x_{3} = 1, \\
x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2 \\
x_{1} x_{2} x_{3} = -1.
\end{array}\right.
$$

Therefore, $a$, $b$, and $c$ are the roots of the equation $x^{3} + 2 x^{2} - x - 1 = 0$; simultaneously, $x_{1}$, $x_{2}$, and $x_{3}$ are the roots of the equation $x^{3} - x^{2} - 2 x + 1 = 0$. If we substitute $x \rightarrow -\frac{1}{x}$ in the equation $x^{3} + 2 x^{2} - x - 1 = 0$, we get the equation $x^{3} - x^{2} - 2 x + 1 = 0$. This is an interesting conclusion.
Using conclusion (6), we can derive many symmetric expressions involving $a$, $b$, and $c$. However, if we need to find the values of $a^{2} b + b^{2} c + c^{2} a$ and $a b^{2} + b c^{2} + c a^{2}$, the above solution is not straightforward. We will use a different approach to find these values.
Clearly, the six roots of these three equations must be the roots of the equation $\left(x^{2} + a x + b\right)\left(x^{2} + b x + c\right)\left(x^{2} + c x + a\right) = 0$, which is $x^{6} + (a + b + c) x^{5} + (a + b + c + b c + c a + a b) x^{4} + (a^{2} + b^{2} + c^{2} + b c + c a + a b + a b c) x^{3} + (a^{2} b + b^{2} c + c^{2} a + b c + c a + a b) x^{2} + (a b^{2} + b c^{2} + c a^{2}) x + a b c = 0$.

Let $s_{1} = x_{1} + x_{2} + x_{3}$,
$s_{2} = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2}$, and $s_{3} = x_{1} x_{2} x_{3}$.
By the relationship between the roots and coefficients, and according to conclusion (6), we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
s_{1} = x_{1} + x_{2} + x_{3} = 1, \\
s_{2} = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2, \\
s_{3} = x_{1} x_{2} x_{3} = -1.
\end{array}\right. \\
\left\{\begin{array}{l}
a + b + c = -2 s_{1}, \\
a + b + c + b c + c a + a b = s_{1}^{2} + 2 s_{2}, \\
a^{2} + b^{2} + c^{2} + b c + c a + a b + a b c \\
= -2 s_{1} s_{2} - 2 s_{3}, \\
a^{2} b + b^{2} c + c^{2} a + b c + c a + a b \\
= s_{2}^{2} + 2 s_{1} s_{3}, \\
a b^{2} + b c^{2} + c a^{2} = -2 s_{2} s_{3}, \\
a b c = s_{3}^{2}.
\end{array}\right.
\end{array}
$$

From equations (7) and (8), we get
$$
\begin{array}{l}
a^{2} b + b^{2} c + c^{2} a = s_{2}^{2} + 2 s_{1} s_{3} - (b c + c a + a b) \\
= (-2)^{2} + 2 \times 1(-1) - (-1) = 3, \\
a b^{2} + b c^{2} + c a^{2} = -2 s_{2} s_{3} \\
= -2(-2)(-1) = -4.
\end{array}
$$