Question

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

Answer 1

Giải:

Ta có: \(VT=a^3+b^3+c^3-3abc\)

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

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

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

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

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

\(=VP\) (Đpcm)

Answer 2

Ta có:

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

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

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

\(\rightarrow\) đpcm

Chúc bạn học tốt!!!

Answer 3

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

Ta có VP: \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(=a^3+ab^2+ac^2-a^2b-abc-a^2c+ba^2+b^3+bc^2-ab^2-b^2c-bac+ca^2+cb^2+c^3-cab-bc^2-ac^2\)

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

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