To solve the cubic equation \( x^3 - 3x^2 + 3x - 1 = 0 \), we can look for rational roots using the Rational Root Theorem, which states that any rational root of the polynomial equation \( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0 \) must be a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \).

In this case, the polynomial is \( x^3 - 3x^2 + 3x - 1 \), where the constant term \( a_0 \) is \(-1\) and the leading coefficient \( a_n \) is \(1\). The possible rational roots are the factors of \(-1\), which are \(\pm 1\).

Let's test these possible roots:

1. Test \( x = 1 \):
\[
1^3 - 3(1)^2 + 3(1) - 1 = 1 - 3 + 3 - 1 = 0
\]
So, \( x = 1 \) is a root of the polynomial.

Since \( x = 1 \) is a root, we can factor the polynomial as \( (x - 1) \) times a quadratic polynomial. To find the quadratic polynomial, we can perform polynomial division or use synthetic division.

Using synthetic division:

\[
\begin{array}{r|rrrr}
1 & 1 & -3 & 3 & -1 \\
& & 1 & -2 & 1 \\
\hline
& 1 & -2 & 1 & 0 \\
\end{array}
\]

The quotient is \( x^2 - 2x + 1 \), so we have:
\[
x^3 - 3x^2 + 3x - 1 = (x - 1)(x^2 - 2x + 1)
\]

Next, we factor the quadratic polynomial \( x^2 - 2x + 1 \):
\[
x^2 - 2x + 1 = (x - 1)^2
\]

So, the original polynomial can be factored as:
\[
x^3 - 3x^2 + 3x - 1 = (x - 1)^3
\]

Setting this equal to zero gives:
\[
(x - 1)^3 = 0
\]

Therefore, the only solution to the equation \( x^3 - 3x^2 + 3x - 1 = 0 \) is:
\[
x = 1
\]

This root has a multiplicity of 3.