jmc

We will use the identity $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz).$$Setting $x=a^2-b^2,$ $y=b^2-c^2,$ $z=c^2-a^2,$ we get $$(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-a^2{)}^3-3(a^2-b^2)(b^2-c^2)(c^2-a^2)=0.$$Setting $x=a-b,$ $y=b-c,$ $z=c-a,$ we get $$(a-b{)}^3+(b-c{)}^3+(c-a{)}^3-3(a-b)(b-c)(c-a)=0.$$Hence, $$\begin{array}{rl}\frac{(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-a^2{)}^3}{(a-b{)}^3+(b-c{)}^3+(c-a{)}^3}& =\frac{3(a^2-b^2)(b^2-c^2)(c^2-a^2)}{3(a-b)(b-c)(c-a)}\\ & =\frac{(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{(a-b)(b-c)(c-a)}\\ & =\boxed{(a+b)(a+c)(b+c)}.\end{array}$$