Firstly try subbing in numbers that are multiples of 27 and see if the function f(x)=0, (1,3,-3,-1,9,-9,27,-27)
Let's try x=3. This gives 27-3(9)-9(3)+27=27-27-27+27=0. Hence the first solution is x=3. We can now use this to find the 3 other solutions. I know there are 3 other solutions as it is a cubic function.
Now as x=3 this gives in the bracket (x-3). Use this to solve to find the quadratic equation.
(x-3)(ax^2+bx+c) = x^3-3x^2-9x+27 where a,b and c are constants.
Lets first find c as this is the easiest.
-3*c=27 so c is -9
Finding a is also easy as it must be 1. x*ax^2=x^3.
Now lets try for the x terms. You can also do x^2 any will work. or expand the brackets and equate co-effecients.
-3ax^2 + bx^2 are the only x^2 terms. -3ax^2 + bx^2=-3x^2 Take out common multiple x^2
-3a+b=-3 we know that a is 1 from before, therefore -3+b=-3 b=0
Now the solutions we have is
(x-3)(x^2-9) Now factorise the quadratic which is a difference of squares so the final solutions:
(x-3)(x+3)(x-3)=0 therefore x=3 or -3(repeated root).
Hope that makes sense if not say and i'll try to explain it more. I'm british so some of my terms i used may be different. :) Good luck.