若方程x^3-x+1=0在(a,b) (a,b是整数,且b-a=1)上有一根,则a+b=?

Answer: a=-2, b=-1, a+b=-3

For an easy way, you can simply use Mathematica to dissolve this equation and get the solution to x^3-x+1=0

Another way is to use the relation of differential
y=x^3-x+1
dy/dx=3x^2-1
when dy/dx=3x^2-1=0, x=1/squart(3) or x=-1/squart(3)
...
You can find that at x=-1, y=1; x=-2, y=-5
Then only solution is between -1 and -2.
So a+b=-3