a3+b3+c3−3abc
=a3+3ab(a+b)+b3+c3−3abc−3ab(a+b)
=(a+b)3+c3−3ab(a+b+c)
=(a+b+c)(a2+2ab+b2−ab−ac+c2)−3ab(a+b+c)
=(a+b+c)(a2+b2+c2−ab−bc−ca)
Ta có : a+b+c=0 => a+b=-c
=> (a+b)3=-c3
=> a3+3a2b+3ab2+c3=-c3
=> a3+b3+(3a2b+3ab2)=-c3
=> a3+b3+3ab(a+b)=-c3
=> a3+b3+3ab(-c)=-c3
=> a3+b3-3abc=-c3
=> a3+b3+c3=3abc
=> a3+b3+c3-3abc=0 (đpcm)
Ta có: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\) (*)
\(\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\).Thế (*) vào ta có:
\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)\(\Leftrightarrow a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)\)
\(\Leftrightarrow a^3+b^3+c^3-3a\left(-b\right)\left(-c\right)=0\) hay
\(a^3+b^3+c^3-3abc=0^{\left(đpcm\right)}\)
