Question

phân tích đa thức thành nhân tử

bc(b+c)+ca(c-a)-ab(a+b)

Answer 1

bc(b+c)+ca(c-a)-ab(a+b)

=bc(b+c)+ c²a- ca²- a²b- ab²

=bc(b+c)- ab(b+c) + ac(b+c) - a²(b+c)

=(b+c)(bc- ab+ ac -a²)

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

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

Answer 2

\(bc\left(b+c\right)+ca\left(c-a\right)-ab\left(a+b\right)\\ =bc\left(b+c\right)+ca\left(c-a\right)-ab\left(a+b\right)-\left(abc-abc\right)\\ =b^2c+bc^2+c^2a-ca^2-a^2b-ab^2-abc+abc\\ =\left(b^2c+bc^2+c^2a+abc\right)-\left(ca^2+a^2b+ab^2+abc\right)\\ =c\left(b^2+bc+ca+ab\right)-a\left(ca+ab+b^2+bc\right)\\ =\left(c-a\right)\left(ca+ab+b^2+bc\right)\\ =\left(c-a\right)\left[\left(ca+ab\right)+\left(b^2+bc\right)\right]\\ =\left(c-a\right)\left[a\left(b+c\right)+b\left(b+c\right)\right]\\ =\left(c-a\right)\left(a+b\right)\left(b+c\right)\)