帮帮忙，解数学题

f'(x)=3x^2-6ax+3b
x=1时,f'(x)=3-6a+3b=-12(直线12x+y-1=0的斜率)
2a-b=5---(1)

把（1，-11）代入f（x)=x^3-3ax^2+3bx得:
1-3a+3b=-11
a-b=4---(2)

解(1)(2)得:
a=1
b=-3

a=1,b=-3

f(x)=x^3-3ax^2+3bx,

so -11=1-3a+3b

tangential line slope 0 -12

f'(x)=3x^2-6ax+3b=-12 when x=1

f′(X)=3x^2-6ax+3b
f（x)的图像与直线12x+y-1=0相切于点（1，-11）
f′(X)=3x^2-6ax+3b=3-6a+3b=-12
-11=1-3a+3b
解得a =1b=-3