帮忙解道数学题，非常急

x^3=a(x-1)^3+b(x-1)^2+c(x-1)+d
=ax^3-3ax^2+3ax-a+bx^2-2bx+b+cx-c+d
=ax^3+(b-3a)x^2+(3a-2b+c)x+(b+d-a-c)
a=1
b-3a=0
3a-2b+c=0
b+d-a-c=0
解得a=1,b=3,c=3,d=1

a(x-1)(x-1)(x-1)=ax^3-3ax^2+3ax-a
b(x-1)(x-1)=bx^2-2bx+b
c(x-1)=cx-c
d=d

a=1
-3a+b=0
3a-2b+c=0
-a+b-c+d=0
得 a=1 b=3 c=3 d=1

x^3 = a(x-1)^3 + b(x-1)^2 + c(x-1) + d
= a x^3 + (b-3a)x^2 + (3a-2b+c)x + b-a+d-c
两边比较系数：
1 = a
0 = b-3a
0 = 3a-2b+c
0 = b-a+d-c
求出：a,b,c,d,= 1,3,3,1