令a+b+c=x=0,ab+bc+ca=y, 则由4=a^2+b^2+c^2=x^2-2y知y=-2 a^4+b^4+c^4=(a^2+b^2+c^2)^2-2[(ab)^2+(bc)^2+(ac)^2] =16-2(y^2-2x*abc) =16-2y^2 =8
