Ta co: \(a^4+b^4+c^4\)
\(=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=\left(a^2+b^2+c^2\right)^2-2\left[\left(ab+bc+ca\right)^2-2\left(ab^2c+abc^2+a^2bc\right)\right]\)
\(=\left(a^2+b^2+c^2\right)^2-2\left[\left(ab+bc+ca\right)^2-2abc\left(a+b+c\right)\right]\)
\(=\left(a^2+b^2+c^2\right)^2-2\left(ab+bc+ca^2\right)\)
\(\Rightarrow2\left(a^4+b^4+c^4\right)=2\left(a^2+b^2+c^2\right)^2-4\left(ab+bc+ca\right)^2\)
\(\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2-\left(2ab+2bc+2ca\right)^2\) \(\left(1\right)\)
Lại có: \(a+b+c=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Rightarrow2ab+2bc+2ca=-\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\left(2ab+2bc+2ca\right)^2=\left(a^2+b^2+c^2\right)^2\) \(\left(2\right)\)
Từ (1) và (2) suy ra
\(2\left(a^4+b^4+c^4\right)=2\left(a^2+b^2+c^2\right)^2-\left(a^2+b^2+c^2\right)^2\)
\(=\left(a^2+b^2+c^2\right)^2\)
