一元一次方程ax^2+bx+c=0(a不等于0)的两根为X1和X2求|X1-X2|，X1^3+X2^

依据求根公式得：x=[-b±√b^2-4ac]/2a
∴|x1-x2|=[√b^2-4ac]/|a|

x1^3+x2^3
=(x1+x2)(x1^2+x2^2-x1x2)
=-(b/a)[(x1+x2)^2-3x1x2]
=-(b/a)[(b/a)^2-3c/a]
=-(b/a)^3+3bc/a^2

x1+x2=-b/a
x1x2=c/a

(x1-x2)^2=(x1+x2)^2-4x1x2=b^2/a^2-4c/a
|x1-x2|=根号(b^2-4ac)/|a|

x1^3+x2^3=(x1+x2)(x1^2+x2^2-x1x2)
=(x1+x2)[(x1+x2)^2-3x1x2]
=-b/a*[b^2/a^2-3c/a]
=-b^3/a^3+3bc/a^2