g(x)=ax³+bx²+cx+d--->f(x)=g'(x)=3ax²+2bx+c
∵a+b+c=0--->c=-(a+b)
∴f(0)f(1)=c(3a+2b+c)=-(a+b)(2a+b)＞0
a＞0--->(1+b/a)(2+b/a)＜0--->-2＜b/a＜-1

(x1-x2)²=(x1+x2)²-4x1x2
　　　　=[(-2b)/(3a)]²-4c/(3a)
　　　　= 4(b²-3ac)/(9a²)
　　　　= 4(b²+3a(a+b)/(9a²)
　　　　= (4/9)(b²+3ab+3a²)/a²
　　　　= (4/9)[(b/a)²+3(b/a)+3]
　　　　= (4/9)[(b/a+3/2)²+(3/4)]
∵-2＜b/a＜-1--->-1/2＜b/a+3/2＜1/2
--->0≤(b/a+3/2)²＜1/4
--->1/3≤(x1-x2)²＜4/9
--->√3/3≤|x1-x2|＜2/3................选A