Lời giải:
Đặt \(\left(\frac{a}{b^2}, \frac{b}{c^2}, \frac{c}{a^2}\right)=(x,y,z)\)
\(\Rightarrow xyz=\frac{abc}{a^2b^2c^2}=\frac{1}{abc}=1\)
Theo bài ra ta có: \(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}=\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\)
\(\Leftrightarrow x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow x+y+z=xy+yz+xz\)
\(\Leftrightarrow (xy-x-y+1)-1+z(x+y-1)=0\)
\(\Leftrightarrow (xy-x-y+1)+z(x+y-1-xy)=0\)
\(\Leftrightarrow (x-1)(y-1)-z(x-1)(y-1)=0\)
\(\Leftrightarrow (x-1)(y-1)(1-z)=0\)
\(\Leftrightarrow \frac{a-b^2}{b^2}.\frac{b-c^2}{c^2}.\frac{a^2-c}{a^2}=0\)
\(\Leftrightarrow (a-b^2)(b-c^2)(c-a^2)=0\)
Do đó ta có đpcm.
