a+b+c=0=>(a+b+c)2=0
=>a2+b2+c2+2ab+2bc+2ca=0
vi a2+b2+c2=1=> 1 + 2(ab+bc+ca)=0=> ab+bc+ca= \(\frac{-1}{2}\)
=>(ab+bc+ca)2=\(\frac{1}{4}\)
=>a2b2+b2c2+c2a2+2abc(a+b+c)=1/4
=>a2b2+b2c2+c2a2=1/4
a2+b2+c2 =1
=> a4+b4+c4+2(a2b2+b2c2+c2a2)=1
=>a4+b4+c4+1/2=1
=>a4+b4+c4=1/2(dpcm)