Given non-zero real numbers $a$, $b$, $c$, satisfying
$$
(a+b+c)(b+c-a)(c+a-b)(a+b-c)=a^{4} \text{. }
$$

Find the maximum and minimum values of $\frac{b}{c}-\frac{c}{b}$, and indicate the values of $|a|:|b|:|c|$ when $\frac{b}{c}-\frac{c}{b}$ takes its maximum and minimum values.
Consider $a$ as the main variable, obtaining a quartic equation in $a$:
$$
2 a^{4}-2\left(b^{2}+c^{2}\right) a^{2}+\left(b^{2}-c^{2}\right)^{2}=0 .
$$

Then, regard equation (1) as a quadratic equation in the unknown $a^{2}$.
Since the equation has real solutions, we have:
$$
\begin{array}{l}
\Delta=4\left(b^{2}+c^{2}\right)^{2}-8\left(b^{2}-c^{2}\right)^{2} \geqslant 0 \\
\Rightarrow\left(b^{2}-c^{2}\right)^{2} \leqslant 4 b^{2} c^{2} \\
\Rightarrow\left(\frac{b}{c}-\frac{c}{b}\right)^{2} \leqslant 4 \\
\Rightarrow-2 \leqslant \frac{b}{c}-\frac{c}{b} \leqslant 2 .
\end{array}
$$

The equality holds if and only if equation (1) has two equal positive real roots, in which case:
$$
\left\{\begin{array} { l } 
{ b ^ { 4 } + c ^ { 4 } = 6 b ^ { 2 } c ^ { 2 } , } \\
{ a ^ { 2 } = \frac { b ^ { 2 } + c ^ { 2 } } { 2 } }
\end{array} \Rightarrow \left\{\begin{array}{l}
b^{2}=\frac{2 \pm \sqrt{2}}{2} a^{2}, \\
c^{2}=\frac{2 \mp \sqrt{2}}{2} a^{2} .
\end{array}\right.\right.
$$

Therefore, when $\frac{b}{c}-\frac{c}{b}$ takes the maximum value 2,
$|a|:|b|:|c|=\sqrt{2}: \sqrt{2+\sqrt{2}}: \sqrt{2-\sqrt{2}}$; when $\frac{b}{c}-\frac{c}{b}$ takes the minimum value -2,
$$
|a|:|b|:|c|=\sqrt{2}: \sqrt{2-\sqrt{2}}: \sqrt{2+\sqrt{2}} \text {. }
$$