已知a+b+c=0,a²+b²+c²=4.
2个回答

由 (a+b+c)^2

=a^2+b^2+c^2+2ab+2bc+2ac

=0

得 ab+bc+ac

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

=-2

同样由

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

=a^4+b^4+c^4+2〔(ab)^2+(bc)^2+(ac)^2〕

=16……………①

又因为a+b+c=0,所以

(ab+bc+ac)^2

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

=(ab)^2+(bc)^2+(ac)^2

=4

即 (ab)^2+(bc)^2+(ac)^2=4……………②

把②式代入①式得到

a^4+b^4+c^4+2〔(ab)^2+(bc)^2+(ac)^2〕

=a^4+b^4+c^4+2*4

=16

所以,a^4+b^4+c^4=8