Question

CMR :

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

Answer 1

Ta sẽ chứng minh \(a^3+b^3+c^3-3abc=0\Leftrightarrow a+b+c=0\)

Phân tích thành nhân tử :  \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

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

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

Vì a + b + c = 0 nên \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) hay \(a^3+b^3+c^3-3abc=0\Leftrightarrow a^3+b^3+c^3=3abc\)

 

 

Answer 2

Ta có : \(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\Leftrightarrow a^3+b^3+3ab\left(a+b\right)+c^3=0\)

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

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