Question

[Nguyễn Thị Huyền]

\(Cho\)\(:\)\(a+b+c=0\)

\(CMR:a^3+b^3+c^3=3abc\)

Answer 1

[Bui Huyen]

\(a+b+c=0\Rightarrow\left(a+b+c\right)^3=0\)

\(\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3=0\)

\(a^3+b^3+3ab\left(a+b\right)+c^3=0\)

\(a^3+b^3+c^3+3ab\left(-c\right)=0\)

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

Answer 2

[★๖ۣۜßảo๖ۣۜPɦα♏๖ۣۜ[EηgĻïšħ☯€lub]★]

Ta có:\(a^3+b^3+c^3-3abc\)

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

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

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

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

\(=0\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\left(dpcm\right)\)