Question

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

Answer 1

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

Ta có : \(VT=a^3+b^3+c^3-3abc\)

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

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

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

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

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

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

\(\left(đpcm\right)\)