Có: \(a^2+b^2+c^2=1\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=1\)
\(\Rightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
Lại có: \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Rightarrow2\left(ab+bc+ac\right)=-1\)
\(\Rightarrow ab+bc+ac=-\frac{1}{2}\)
\(\Rightarrow\left(ab+bc+ac\right)^2=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=\frac{1}{4}-2abc\left(a+b+c\right)\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=\frac{1}{4}\)
Vậy: \(a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4=1-2.\frac{1}{4}=1-\frac{1}{2}=\frac{1}{2}\)
