Từ \(a+b+c=0\) suy ra \(\begin{cases}c=-\left(a+b\right)\\a+b=-c\end{cases}\) thì:
\(a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3\)
\(=a^3+b^3-\left[a^3+3ab\left(a+b\right)+b^3\right]\)
\(=a^3+b^3-a^3-3ab\left(a+b\right)-b^3\)
\(=-3ab\left(a+b\right)=-3ab\left(-c\right)\)
\(=3abc=3\cdot11=33\)