Question

Phân tích đa thức thành nhân tử:   a³+b³+c³-3abc

Answer 1

Ta Có : \(a^3+b^3+c^3-3abc \)

\(=\left(a+b\right)\left(a^2+ab+b^2\right)+c\left(c^2-3ab\right)\)

\(=a^3+3a^2b+ab^2+b^3+c^3-3abc-3a^2b-3ab^2\)

\(=\left(a+b\right)^3+c^3-3abc\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

= \(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

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