A(b-c)³+b(c-a)³+c(a-b)³

Đáp án:

 $a.(b-c)^3+b.(c-a)^3+c.(a-b)^3$

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

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

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

$= ab^3-ac^3+bc^3-b^3c-a^3b+a^3c$