Question

a³+b³-c³+3abc

phân tích đa thức thành nhân tử

Answer 1

a³+3a²b+3ab²+b³-c³+3abc-3a²b-3ab²

= (a³+3a²b+3ab²+b³)-c³-(3a²b+3ab²-3abc)

= (a+b)³-c³-3ab(a+b-c)

=(a+b-c)[(a+b)²+c(a+b)+c²]-3ab(a+b-c)

=(a+b-c)(a²+2ab+b²+ac+cb+c²-3ab)

=(a+b-c)(a²+b²+c²-ab+ac+cb)

Answer 2

\(a^3+b^3-c^3+3abc=\left(a+b\right)^3-3ab\left(a+b\right)-c^3+3abc\\ =\left(a+b-c\right)^3-3c\left(a+b\right)\left(a+b-c\right)-3ab\left(a+b-c\right)\\ =\left(a+b-c\right)\left[\left(a+b-c\right)^2-3ac-3bc-3ab\right]\\ =\left(a+b-c\right)\left(a^2+b^2+c^2-ab-5bc-5ac\right)\)