Ta có : \(a+b+c=0\Leftrightarrow b+c=-a\)
\(\Rightarrow\left(b+c\right)^2=a^2\)(1)
\(\Rightarrow\left(a^2-b^2-c^2\right)^2=4b^2c^2\)
\(\Leftrightarrow a^4+b^4+c^4-2\left(a^2b^2-b^2c^2+2c^2a^2\right)=4b^2c^2\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
Từ (1) ta có :
\(a^4+b^4+c^4=2\left(ab+bc+ac\right)^2-4abc\left(a+b+c\right)\)
\(=2\left(ab+bc+ca\right)^2\)
Vì a + b + c = 0
Ta có đpcm
+) a^4 + b^4 + c^4 = ( a + b + c ) ^4
= 0^4 =0
+) 2( ab + bc + ca ) ^2 = 2( abc (a + b +c ))^2
=2(abc*0)^2
=0
vậy a^4+b^4+c^4=2(ab+bc+ca)^2(=0)
\(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2ab-2bc-2ac\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left(2ab+2bc+2ac\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=4\left(ab+bc+ac\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4=4\left(ab+bc+ac\right)^2-2a^2b^2-2a^2c^2-2b^2c^2\)
Ta có : \(2\left(ab+bc+ac\right)^2-2a^2b^2-2a^2c^2-2b^2c^2\)
\(=2\left(a^2b^2+a^2c^2+b^2c^2+2ab^2c+2a^2bc+2abc^2\right)-2a^2b^2-2a^2c^2-2b^2c^2\)
\(=-2abc\left(a+b+c\right)=0\)
Do đó \(4\left(ab+bc+ac\right)^2-2a^2b^2-2a^2c^2-2b^2c^2=2\left(ab+bc+ac\right)^2\)
Hay \(a^4+b^4+c^4=2\left(ab+bc+ac\right)^2\) (đpcm)
