已知a+b+c＝0，a＾2+b＾2+c＾2＝4求A＾4＋B＾4+C＾4

(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2*b^2+2b^2*c^2+2c^2*a^2；
其中2a^2*b^2+2b^2*c^2+2c^2*a^2
=a^2*(b^2+c^2)+b^2*(c^2+a^2)+c^2*(a^2+b^2)
=a^2*((b+c)^2-2bc)+b^2*((c+a)^2-2ac)+c^2*((a+b)^2-2ab)
=a^2*((-a)^2-2bc)+b^2*((-b)^2-2ac)+c^2*((-c)^2-2ab)
=a^4+b^4+c^4-2abc*(a+b+c)
=a^4+b^4+c^4；
所以(a^2+b^2+c^2)^2=2(a^4+b^4+c^4)

a^4+b^4+c^4=1/2(a^2+b^2+c^2)^2=16/2=8

（a^2+b^2+c^2）^2=a^4+b^4+c^4+2a^2b^2+2a^2b^2+2b^2c^2=4^2=16 ①
(a+b+c)^2=a^2+b^2+c^2+2a^2b^2+2a^2c^2+2b^2c^2
a^2+b^2+c^2=-(2a^2b^2+2a^2b^2+2b^2c^2)
( a^2+b^2+c^2)^2= (2a^2b^2+2a^2b^2+2b^2c^2)^2
a^4+b^4+c^4+2a^2b^2+2a^2b^2+2b^2c^2=4a^2b^2+4a^2b^2+4b^2c^2+4abc(a+b+c)=0^2=0
a^4+b^4+c^4-( 2a^2b^2+2a^2b^2+2b^2c^2)=0 ②
①+②=2(a^4+b^4+c^4)=16
a^4+b^4+c^4=8