This is a typical symmetric polynomial factorization problem. Let $f(a, b, c)=(ab+bc+ca)(a+b+c)-abc+(a+b)(b+c)(c+a)$. When $a=-b$, $f(a,b,c)$ will be $0$. Therefore $(a+b)|f(a,b,c)$. By symmetry, it must hold that $$f(a, b, c)= k(a+b)(b+c)(c+a)$$ Setting $a=b=c=1 \implies k=2$. Therefore $$f(a,b,c)=\boxed{2(a+b)(b+c)(c+a)}$$