Từ giả thiết: \(a+b+c=0\Leftrightarrow c=-\left(a+b\right)\). Ta có:
\(a^3+b^3+c^3=a^3+b^3+\left[-\left(a+b\right)\right]^3\)
\(=a^3+b^3-\left(a+b\right)^3\)
\(=a^3+b^3-\left[a^3+b^3+3ab\left(a+b\right)\right]\)
\(=a^3+b^3-a^3-b^3-3ab\left(a+b\right)\)
\(=3ab\left[-\left(a+b\right)\right]\)
\(=3abc\left(đpcm\right)\)
Ta có : a + b + c = 0
=> a + b = - c
=> a3 + b3 + 3ab( a + b ) = ( - c)3
=> a3 + b3 + c3 = -3ab( a + b )
= 3abc
