<s>[INST] $bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$ 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.
 [/INST] $bc(b-c)+ca(c-a)+ab(a-b)=c[b(b-c)+a(c-a)]+ab(a-b)=c[b^2-bc+ac-a^2]+ab(a-b)=c[b^2-a^2+ac-bc]+ab(a-b)=c[(b-a)(b+a)-c(b-a)]-ab(b-a)=c(b-a)(b+a-c)-ab(b-a)=(b-a)[c(b+a-c)-ab]=-(a-b)(bc+ac-c^2-ab)=-(a-b)[c(b-c)-a(b-c)]=-(a-b)(c-a)(b-c)$
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