nếu sửa lại a+b+c=0 thì
a+b+c=0
=>02=(a+b+c)2
<=>0=a2+b2+c2+2ab+2bc+2ac
<=>0=1+2.(ab+bc+ac)
<=>ab+bc+ac=-1/2
=>(ab+bc+ac)2=1/4
<=>a2b2+b2c2+a2c2+2ab2c+2a2bc+2abc2=1/4
<=>a2b2+b2c2+a2c2+2abc(a+b+c)=1/4
<=>a2b2+b2c2+a2c2+2abc.0=1/4
<=>a2b2+b2c2+a2c2=1/4
ta có:
(a2+b2+c2)2=a4+b4+c4+2a2b2+2b2c2+2a2c2
<=>12=a4+b4+c4+2.(a2b2+b2c2+a2c2)
<=>1=a4+b4+c4+2.1/4
<=>a4+b4+c4=1/2