Question

Phân tích đa thức sau thành nhân tử a(b+c)^2(b-c) +b(c+a)^2(c-a) +c(a+b)^2(a-b)

Answer 1

Mình không biết

Answer 2

ko bt thì  ko nói nha mình đang cần gấp lém xin đừng trêu

Answer 3

\(a\left(b+c\right)^2\left(b-c\right)+b\left(c+a\right)^2\left(c-a\right)+c\left(a+b\right)^2\left(a-b\right)\)

\(=a\left(b+c\right)^2-\left(b-c\right)-b\left(c+a\right)^2\left(b-c+a-b\right)+c\left(a+b\right)^2\left(a-b\right)\)

\(=\left(b-c\right)\left[a\left(b+c\right)^2-b\left(c+a\right)^2\right]-\left(a-b\right)\left[b\left(c+a\right)^2-c\left(a+b\right)^2\right]\)

\(=\left(b-c\right)\left(ab^2+ac^2-bc^2-ba^2\right)-\left(a-b\right)\left(bc^2+ba^2-ca^2-cb^2\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(c^2-ab\right)-\left(a-b\right)\left(b-c\right)\left(a^2-bc\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(c^2-ab-a^2+bc\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)