Factor the expression A3(B-C) + B3(C-A) + C3(A-B)?

Hint 1





Note that the expression is zero if B = C, or if A = C, or if A = B. This means that B - C, C - A, A - B must all be factors. If you factor these out, what is left over?


Hint 2





There are a couple of ways to find the remaining factor. The brute-force method is to multiply out the original expression, use polynomial long division to take out the known factors, then see what is left over. But there is another way you can reason out the remaining factor. Look at the symmetries in the original expression and in the known factors; what do they tell us about the remaining factors?


Hint 3





The original expression has a circular symmetry in A, B, C; that is, if you substitute A for B, B for C, and C for A, you get the same expression over again. If you make this substitution in the known factors (B-C)(C-A)(A-B) you also get the same expression over again. Therefore what?

Factor the expression A3(B-C) + B3(C-A) + C3(A-B)?
Given expression


=a^3(b-c) +b^3 c-b^3 a+c^3 a-c^3 b


=a^3(b-c) -(ab^3-ac^3)+(b^3 c-bc^3)


=a^3(b-c)-a(b^3-c^3)+bc(b^2-c^2)


=a^3(b-c)-a(b-c)(b^2+bc+c^2)+bc(b+c)(b...


=(b-c){a^3-a(b^2+bc+c^2)+bc(b+c)


=(b-c)(a^3-ab^2-abc-ac^2+b^2c+bc^2)


=(b-c)(b^2c-ab^2+bc^2-abc-ac^2+a^3)


=(b-c){b^2(c-a)+bc(c-a)-a(c^2-a^2)


=(b-c)(c-a){b^2+bc-a(c+a)}


=(b-c)(c-a)(b^2+bc-ac-a^2)


=(b-c)(c-a)(bc-ac+b^2-a^2)


=(b-c)(c-a){c(b-a)+(b+a)(b-a)}


=(b-c)(c-a)(b-a)(c+a+b)


= -(a-b)(b-c)(c-a)(a+b+c)
Reply:A3(B-C)+B3(C-A)+C3(A-B)


=AB3-AC3+BC3-AB3+AC3-BC3


= 0 Answer