To solve the equation x^3 - 9x^2 + 27x - 27 = 0, we can use the Rational Root Theorem to find possible rational roots.

The Rational Root Theorem states that if a polynomial equation has a rational root p/q, where p is a factor of the constant term and q is a factor of the leading coefficient, then p/q is a possible rational root.

In this case, the constant term is -27 and the leading coefficient is 1. The factors of -27 are ±1, ±3, ±9, ±27, and the factors of 1 are ±1.

Therefore, the possible rational roots are ±1, ±3, ±9, ±27.

We can now test these possible roots by substituting them into the equation to see if they satisfy the equation.

When we substitute x = 1 into the equation, we get:
1^3 - 9(1)^2 + 27(1) - 27 = 1 - 9 + 27 - 27 = -8 + 27 - 27 = -8
Since -8 is not equal to 0, x = 1 is not a root.

When we substitute x = -1 into the equation, we get:
(-1)^3 - 9(-1)^2 + 27(-1) - 27 = -1 - 9 - 27 - 27 = -1 - 9 - 54 = -64
Since -64 is not equal to 0, x = -1 is not a root.

When we substitute x = 3 into the equation, we get:
3^3 - 9(3)^2 + 27(3) - 27 = 27 - 81 + 81 - 27 = 27 - 27 = 0
Since 0 is equal to 0, x = 3 is a root.

Therefore, x = 3 is a root of the equation x^3 - 9x^2 + 27x - 27 = 0.

To find the other roots, we can use polynomial long division or synthetic division to divide the polynomial by (x - 3).

Using synthetic division, we have:
3 | 1 - 9 27 - 27
| 3 -18 27
|________________
1 - 6 9 0

The quotient is x^2 - 6x + 9.

Now we have a quadratic equation x^2 - 6x + 9 = 0, which can be factored as (x - 3)(x - 3) = 0.

Therefore, the equation x^3 - 9x^2 + 27x - 27 = 0 can be factored as (x - 3)(x - 3)(x - 3) = 0.

The solutions to the equation are x = 3, x = 3, and x = 3.