Question

Cmr:

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

Answer 1

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

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

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

Answer 2

Ta có : (a+b+c)(a²+b²+c²-ab-bc-ca)

=a³+ab²+ac²-a²b-abc-ca²+a²b+b³+bc²-ab²-b²c-abc+a²c+cb²+c³-abc-bc²-c²a

Trừ đi các hạng tử đồng dạng ta có kết quả :

=a³+b³+c³-3abc

Vậy : a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)