Question

数学题：若a+b+c=0,a^2+b^2+c^2=1 ,求a^4+b^4+c^4的值.

Answer 1

（a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)

0=1+2(ab+ac+bc)

bc+ac+ab=-1/2

2.

(bc+ac+ab)^2=b^2c^2+a^2c^2+a^2b^2+2(abc^2+acb^2+bca^2)

(-1/2)^2=b^2c^2+a^2c^2+a^2b^2+2abc(a+b+c)

1/4=b^2c^2+a^2c^2+a^2b^2+0

b^2c^2+a^2c^2+a^2b^2=1/4

a^4+b^4+c^4

=（a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)

=1-2(b^2c^2+a^2c^2+a^2b^2

=1-2*1/4

=1-1/2

=1/2