Question

若a+b+c=0,a^2+b^2+c^2=1,试求下列各式的值

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