Question

若a+b+c=0,a^2+b^2+c^2=1,试求下列各式的值

(1)bc+ac+ab
(2)a^4+b^4+c^4

谢谢了,好的追100分(步骤能看懂)

Answer 1

1）
(a+b+c)^2
=a^2+b^2+c^2+2ab+abc+2ca
=1+2(ab+bc+ca)
=0
ab+bc+ca=-1/2

2)
a^4+b^4+c^4
=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)

=1-2[(ab+bc+ca)^2-2(abbc+bcca+caab)]
=1-2[1/4-2abc(b+c+a)]
=1-1/2-4abc*0
=1/2

Answer 2

1、设bc+ac+ab=x
2x+2=a^2+b^2+2ab+c^2+a^2+2ac+b^2+c^2+2bc
=（a+b）^2+(b+c)^2+(c+a)^2
=(-c)^2+(-a)^2+(-b)^2
=1
x=-1/2