You can see that x-3 works if you set the the polynomial equal to zero and let x=3. Then you can divide x^3-27 by x-3 to get the remaining polynomial which is x^2+3x+9, which is unfactorable unless you want to work with complex numbers. So (x-3)(x^2+3x+9).
Comments
(X-3)(x^2+3x+9)
You can see that x-3 works if you set the the polynomial equal to zero and let x=3. Then you can divide x^3-27 by x-3 to get the remaining polynomial which is x^2+3x+9, which is unfactorable unless you want to work with complex numbers. So (x-3)(x^2+3x+9).
Since x=3 is a solution, (x-3) is a factor. I would use synthetic division to figure out the rest: (x-3)(x^2+3x+9).
One roor is 3 so you should find the other factor dividing by
(x-3)
x^3-27 = (x-3)*(x^2+3x+9)
(x-3)(x-3)(x-3)
use difference of cubes.
Easy as Pi.