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, find the value of $a^{2} + b^{2} + c^{2}$.
(The 4th Chen Shengshen Cup National High School Mathematics Olympiad)
Reference [1] provides a detailed solution to this problem. Through further research, it is found that a simpler solution can also be given to find the values of
$$
a^{2} + b^{2} + c^{2}, \quad a^{2} b + b^{2} c + c^{2} a, \quad a b^{2} + b c^{2} + c a^{2}
$$
From [1] we know
$$
x_{1}=\frac{a-b}{a-c}, x_{2}=\frac{b-c}{b-a}, x_{3}=\frac{c-a}{c-b},
$$

where $x_{1}$ is the common root of equations (1) and (3), $x_{2}$ is the common root of equations (1) and (2), and $x_{3}$ is the common root of equations (2) and (3).
Since $x_{1}$ and $x_{2}$ are the two roots of equation (1),
$\Rightarrow x_{1} x_{2}=b=-\frac{b-c}{a-c} \Rightarrow \frac{c-a}{c-b}=-\frac{1}{b}=x_{3}$.
Similarly, $x_{1}=-\frac{1}{c}, x_{2}=-\frac{1}{a}$, and
$x_{1} x_{2}=\frac{1}{a c}=b \Rightarrow a b c=1$.
Since $x_{2}=-\frac{1}{a}$ is a root of equation (1),
$\Rightarrow \frac{1}{a^{2}}-1+b=0 \Rightarrow \frac{1}{a^{2}}=1-b$.
Similarly, $\frac{1}{b^{2}}=1-c, \frac{1}{c^{2}}=1-a$.
Thus, $\frac{1}{a^{2}} \cdot \frac{1}{b^{2}} \cdot \frac{1}{c^{2}}=(1-b)(1-c)(1-a)$.
Combining with equation (4) we get
$1=1-a-b-c+a b+b c+c a-a b c$
$\Rightarrow a b+b c+c a=1+a+b+c$.
Since $x_{2}=-\frac{1}{a}$ is a root of equation (2), we have
$\frac{1}{a^{2}}-\frac{b}{a}+c=0 \Rightarrow 1-a b+a^{2} c=0$.
Combining with equation (4) we get
$a b c-a b+a^{2} c=0(a \neq 0)$
$\Rightarrow b c-b+c a=0$.
Similarly, $a b-a+b c=0, c a-c+a b=0$.
Adding these three equations, we get
$2(a b+b c+c a)=a+b+c$.
Combining equations (5) and (6) we get
$a+b+c=-2, a b+b c+c a=-1$.
Then $a^{2}+b^{2}+c^{2}$
$=(a+b+c)^{2}-2(a b+b c+c a)=6$.
Since $x_{1}=-\frac{1}{c}$ is a root of equation (3),
$\Rightarrow \frac{1}{c^{2}}-1+a=0 \Rightarrow 1-c^{2}+c^{2} a=0$.
Similarly, $1-b^{2}+b^{2} c=0,1-a^{2}+a^{2} b=0$.
Adding these three equations, we get
$a^{2} b+b^{2} c+c^{2} a=a^{2}+b^{2}+c^{2}-3=3$.
Since $x_{3}=-\frac{1}{b}$ is a root of equation (3), we have
$$
\frac{1}{b^{2}}-\frac{c}{b}+a=0 \Rightarrow 1-b c+b^{2} a=0 .
$$

Similarly, $1-a b+a^{2} c=0,1-c a+c^{2} b=0$.
Adding these three equations, we get
$b^{2} a+a^{2} c+c^{2} b=a b+b c+c a-3=-4$.