Nice problem?

it's the same as $ab(b-c)=c(c-a)$ so if $a=c$ then $a=b=c$ suppose they are distinct if $a=0$ then $a=b=c=0$ so by multiplying we find $abc=1$ now notice that $ab^2+ac=c^2+abc=c^2+1 $ $ RHS>1>0 $ so if $a<0$ then $ c<0$ then $ b<0 $ but $ abc =1 $ contradiction so $a,b,c>0$ $WLOG a>b>c$ now from $ab(b-c)=c(c-a)$ $RHS<0 LHS>0$ so contradiction solutions are $a=b=c$ im not sure if its correct