已知 $abc = - 1$，$\dfrac{{{a^2}}}{c} + \dfrac{b}{{{c^2}}} = 1$，${a^2}b + {b^2}c + {c^2}a = t$，求 $a{b^5} + b{c^5} + c{a^5}$ 的值．
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【出处】
2013年清华大学保送生试题
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【答案】
$3$
【解析】
令 $a = - \dfrac{x}{y}$，$b = - \dfrac{y}{z}$，$c = - \dfrac{z}{x}$，则 $\dfrac{{{a^2}}}{c} + \dfrac{b}{{{c^2}}} = 1$ 即$$\dfrac{{{x^2}}}{{{y^2}}} \cdot \left( { - \dfrac{x}{z}} \right) + \left( { - \dfrac{y}{z}} \right) \cdot \left( {\dfrac{{{x^2}}}{{{z^2}}}} \right) = 1,$$整理得 ${x^3}{z^2} + {y^3}{x^2} + {z^3}{y^2} = 0$，于是$${x^6}{y^9} + {y^6}{z^9} + {z^6}{x^9} = 3\left( {{x^2}{y^3}} \right) \cdot \left( {{y^2}{z^3}} \right) \cdot \left( {{z^2}{x^3}} \right)$$因此\[\begin{split} ab^5+bc^5+ca^5 &= \dfrac{x}{y} \cdot \dfrac{y^5}{z^5} + \dfrac{y}{z} \cdot \dfrac{z^5}{x^5} + \dfrac{z}{x} \cdot \dfrac{x^5}{y^5}\\ & = \dfrac{{x{y^4}}}{{{z^5}}} + \dfrac{yz^4}{x^5}+ \dfrac{{z{x^4}}}{{{y^5}}}\\& = \dfrac{x^6y^9 + y^6z^9 + z^6x^9}{x^5y^5z^5} \\&= \dfrac{3\left(x^2y^3\right) \cdot \left(y^2z^3\right) \cdot \left(z^2x^3\right)}{x^5y^5z^5}\\&= 3.\end{split}\]
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