則ax02+bx0+c=0,bx02+cx0+a=0,cx02+ax0+b=0.
把上面三個(gè)式子相加,并整理得
(a+b+c)(x02+x0+1)=0.
因?yàn)?span>
x | 20 |
1 |
2 |
3 |
4 |
所以a+b+c=0.
于是
a2 |
bc |
b2 |
ca |
c2 |
ab |
a3+b3+c3 |
abc |
a3+b3?(a+b)3 |
abc |
?3ab(a+b) |
abc |
故本題選D.
a2 |
bc |
b2 |
ca |
c2 |
ab |
x | 20 |
1 |
2 |
3 |
4 |
a2 |
bc |
b2 |
ca |
c2 |
ab |
a3+b3+c3 |
abc |
a3+b3?(a+b)3 |
abc |
?3ab(a+b) |
abc |