Đặt: $x=\dfrac{a}{b^2};y=\dfrac{b}{c^2};z=\dfrac{c}{a^2} \Rightarrow xyz=1$
Theo bài ra ta có:
$x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$
$\Leftrightarrow x+y+z=xy+yz+zx$
$\Leftrightarrow xyz-(xy+yz+zx)+(x+y+z)-1=0$
$\Leftrightarrow (x-1)(y-1)(z-1)=0$
$\Leftrightarrow \left[\begin{array}{l}x=1\\y=1\\z=1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}a=b^2\\b=c^2\\c=a^2\end{array}\right.$, đpcm.