phân tích đa thức thành nhân tử ạ
$B=(b-c)(b+c)^4+(c-a)(c+a)^4+(a-b)(a+b)^4$
$=(b-a+a-c)(b+c)^4+(c-a)(c+a)^4+(a-b)(a+b)^4$
$=(a-b)[(a+b)^4-(b+c)^4]+(c-a)[(c+a)^4-(b+c)^4]$
$=(a-b)[(a+b)^2-(b+c)^2][(a+b)^2+(b+c)^2]+(c-a)[(c+a)^2-(b+c)^2][(c+a)^2+(b+c)^2]$
$=(a-b)(a-c)(a+2b+c)[(a+b)^2+(b+c)^2]-(a-c)(a-b)(a+b+2c)[(c+a)^2+(b+c)^2]$
$=(a-b)(a-c)[(a+b)^3+(b+c)(a+b)^2+(b+c)^3+(a+b)(b+c)^2-(a+c)^3-(b+c)(c+a)^2-(b+c)^3-(a+c)(b+c)^2]$
$=(a-b)(a-c)[(a+b)^3+(b+c)(a+b)^2+(b-c)(b+c)^2-(a+c)^3-(b+c)(c+a)^2]$
$=(a-b)(a-c)\{(b-c)[(a+b)^2+(a+b)(a+c)+(a+c)^2]+(b+c)(b-c)(2a+b+c)+(b-c)(b+c)^2\}$
$=(a-b)(a-c)(b-c)[(a+b)^2+(a+b)(a+c)+(a+c)^2+(b+c)(2a+b+c)+(b+c)^2]$
$=(a-b)(a-c)(b-c)[(a+b)^2+(b+c)^2+(a+c)^2+(a+b)(b+c)+(a+c)(a+b)+(c+a)(c+b)]$
Có gì khúc mắc em hỏi lại nhé