Cho thêm cái \(a^2+b^2+c^2=1\) là ez rồi
\(a+b+c=1\Leftrightarrow\left[\left(a+b\right)+c\right]^3=1\)
\(\Leftrightarrow\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3=1\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3+3\left(a^2+2ab+b^2\right)c+3ac^2+3bc^2+c^3=1\)
\(\Leftrightarrow\left(a^3+b^3+c^3\right)+3\left(a^2b+ab^2+a^2c+2abc+b^2c+ac^2+bc^2\right)=1\)
\(\Leftrightarrow a^2b+ab^2+a^2c+abc+abc+b^2c+ac^2+bc^2=0\)
\(\Leftrightarrow ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(ab+ac+bc+c^2\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
Xét \(a=-b\). Ta có theo đề bài \(a+b+c=1\Leftrightarrow c=1\)
\(a=-b\Leftrightarrow a^{1981}=-b^{1981}\)
\(S=a^{1981}+b^{1981}+c^{1981}=c^{1981}=1^{1981}=1\)
Các bạn giúp mình nha.Mình gấp lắm.