Question

Phân tích thành nhân tử:
a. a³ + b³ + c³ – 3abc
b. (x - y)³ + (y - z)³ + (z - x)³

Answer 1

a, \(a^3+b^3+c^3-3abc\)

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

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

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

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

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

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

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

b,