Có: \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2+2\left(ab+ac+bc\right)=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+ac+bc\right)\)
Thay: \(a^2+b^2+c^2=1\)
\(\Rightarrow-2\left(ab+ac+bc\right)=1\Rightarrow ab+ac+bc=-\frac{1}{2}\)
Lại có: \(a^2+b^2+c^2=1\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1\)
Mà: \(2a^2b^2+2a^2c^2+2b^2c^2=2\left(a^2b^2+a^2c^2+b^2c^2\right)=2\left(ab+ac+bc\right)^2=2.\left(-\frac{1}{2}\right)^2=\frac{1}{2}\)
\(\Rightarrow a^4+b^4+c^4=1-\left(2a^2b^2+2a^2c^2+2b^2c^2\right)=1-2\left(a^2b^2+a^2c^2+b^2c^2\right)=1-\frac{1}{2}=\frac{1}{2}\)
Vậy: \(a^4+b^4+c^4=\frac{1}{2}\)
