f(x)=x^3+bx^2-9x+c; f(1)=1+b-9+c. We are not given the value of f(1) so let's start by assuming f(1)=0 as a slight addition to the original question. This means that 1+b-9+c=0, b+c=8. f(1)=0 means that 1 is a zero of the expression and x^3+bx^2-9x+c=(x-1)(x^2+ax-c) where a needs to be found. This expands to: x^3+(a-1)x^2-(c+a)x+c so we can write b=a-1 and c+a=9. The quadratic x^2+ax-c can be written x^2+(9-c)x-c, so x=(c-9+sqrt(81-18c+c^2+4c))/2=(c-9+sqrt(c^2-14c+81))/2. The question offers no more information to determine unique values for b and c, but we can put in some constraints because b and c must be rational. c^2-14c+81=n^2, a perfect square.
Completing the square for the left-hand side: (c^2-14c+49)-49+81=n^2; (c-7)^2=n^2-32; c-7=+sqrt(n^2-32). When n=6, c=7+2, so c=9 or 5 and b=-1 or 3, because b+c=8. Therefore f(x)=x^3-x^2-9x+9=(x-1)(x^2-9)=(x-1)(x-3)(x+3) or x^3+3x^2-9x+5=(x-1)(x^2+4x-5)=(x+5)(x-1)^2. Therefore the solutions are (b,c)=(-1,9) or (3,5).