Answer to Question #100874 in Algebra for Sidan

Question

If a,b,c are real numbers and a+b+c=0, What's a(b-c)^3 + b(c-a)^3 + c(a-b)^3 ?

Answer 1

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

$= a(b^3−3b^2c+3bc^2−c^3)+b(c^3−3c^2a+3ca^2−a^3)+c(a^3−3a^2b+3ab^2−b^3)=$

$=ab ^3 −3ab ^2 c+3abc ^2 −ac ^3 + bc ^3 −3abc ^2 +3a ^2 bc−ba ^3 +ca ^3 −3a ^2 bc+3ab ^2 c−cb ^3 =a b ^3 −ac ^3 +bc ^3 −ba ^3 +ca ^3 −cb ^3$

$a+b+c=0⇒c=−a−b$

$ab ^3 −ac ^3 +bc ^3 −ba ^3 +ca ^3 −cb ^3=ab ^3 −ba ^3 +(−a+b)c ^3 +c(a ^3 −b ^3 )=$

$=a b ^3 −ba ^3 +(−a+b)(−a−b) ^3 +(−a−b)(a ^3 −b ^3 )=$

$= ab ^3 −ba ^3 +(−a+b)(−a ^3 −3a ^2 b−3ab ^2 −b ^3 )−a ^4 +ab ^3 −ba ^3 +b ^4=$

$=-a ^4 +2ab ^3 −2ba ^3 +b ^4 +(a ^4 +3a ^3 b+3a ^2 b ^2 +ab ^3 −ba ^3 −3a ^2 b^2 −3ab ^3 −b ^4 )=0$

$a(b−c) ^3 +b(c−a) ^3 +c(a−b) ^3 =0$