Question 3 Given $a b c=-1, a^{2} b+b^{2} c+c^{2} a=t$, $\frac{a^{2}}{c}+\frac{b}{c^{2}}=1$. Try to find the value of $a b^{5}+b c^{5}+c a^{5}$.
【Analysis】Using $a b c=-1$ can make common substitutions
$$
a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x} \text {. }
$$

This problem involves three letters, and can also be solved by the method of elimination.
Solution 1 Let $a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x}$.
$$
\begin{array}{l}
\text { Then } \frac{a^{2}}{c}+\frac{b}{c^{2}}=1 \text { becomes } \\
x^{2} y^{3}+y^{2} z^{3}+z^{2} x^{3}=0 . \\
\text { Hence } a b^{5}+b c^{5}+c a^{5}-3 \\
=\frac{x y^{4}}{z^{5}}+\frac{y z^{4}}{x^{5}}+\frac{z x^{4}}{y^{5}}-3 \\
=\frac{x^{6} y^{9}+y^{6} z^{9}+z^{6} x^{9}-3 x^{5} y^{5} z^{5}}{x^{5} y^{5} z^{5}} \\
=\frac{\left(x^{2} y^{3}+y^{2} z^{3}+z^{2} x^{3}\right)\left(x^{4} y^{6}+y^{4} z^{6}+z^{4} x^{6}-x^{2} y^{5} z^{3}-y^{2} z^{5} x^{3}-z^{2} x^{5} y^{3}\right)}{x^{5} y^{5}} \\
=0 \\
\Rightarrow a b^{5}+b c^{5}+c a^{5}=3 .
\end{array}
$$

Solution 2 From $a b c=-1 \Rightarrow b=-\frac{1}{a c}$.
Substitute into $\frac{a^{2}}{c}+\frac{b}{c^{2}}=1$, we get $a^{3} c^{2}=a c^{3}+1$.
$$
\begin{array}{l}
\text { Then } a b^{5}+b c^{5}+c a^{5} \\
=-\frac{1}{a^{4} c^{5}}-\frac{c^{4}}{a}+c a^{5}=\frac{a^{9} c^{6}-1-a^{3} c^{9}}{a^{4} c^{5}} \\
=\frac{\left(a c^{3}+1\right)^{3}-1-a^{3} c^{9}}{a^{4} c^{5}}=\frac{3\left(a^{2} c^{6}+a c^{3}\right)}{a^{4} c^{5}} \\
=\frac{3\left(a c^{3}+1\right)}{a^{3} c^{2}}=3 .
\end{array}
$$