$bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$

Question

My book has used the equality $$bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$$ But its proof is not given.

When I open the LHS, I get $b^2c-bc^2+c^2a-ca^2+a^2b-ab^2$. I do not know how to proceed next.

Answer 1

$$bc(b-c)+ca(c-a)+ab(a-b)=b^2c-a^2c-bc^2+ac^2+ab(a-b)=$$ $$=c^2(a-b)-c(a-b)(a+b)+ab(a-b)=(a-b)(c^2-ac-bc+ab)=$$ $$=(a-b)(c-a)(c-b).$$

Answer 2

$bc(b-c)+ca(c-a)+ab(a-b)=c[b(b-c)+a(c-a)]+ab(a-b)=c[b^2-bc+ac-a^2]+ab(a-b)=c[b^2-a^2+ac-bc]+ab(a-b)=c[(b-a)(b+a)-c(b-a)]-ab(b-a)=c(b-a)(b+a-c)-ab(b-a)=(b-a)[c(b+a-c)-ab]=-(a-b)(bc+ac-c^2-ab)=-(a-b)[c(b-c)-a(b-c)]=-(a-b)(c-a)(b-c)$