a+b+c=0<=> $a^2+b^2+c^2+2ab+2bc+2ca=0<=>ab+bc+ca=\frac{-1}{2}$ (vì $a^2+b^2+c^2=1$)
$a^2+b^2+c^2=1<=>a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=1<=>a^4+b^4+c^4+2[(ab+bc+ca)^2-2abc(a+b+c)]=1<=>a^4+b^4+c^4+\frac{1}{2}=1=> a^4+b^4+c^4=\frac{1}{2}$.
VẬY A=$\frac{1}{2}$.