The expression you've provided is a polynomial: \(-x^3 + 3x^2 - 3x + 1\).

To analyze this polynomial, we can look at several aspects:

### 1. **Degree and Leading Coefficient:**
The degree of the polynomial is 3 (the highest power of \(x\)), and the leading coefficient (the coefficient of the term with the highest power) is \(-1\).

### 2. **Roots:**
To find the roots of the polynomial, we need to solve the equation \(-x^3 + 3x^2 - 3x + 1 = 0\).

One way to find the roots is to use the Rational Root Theorem, which suggests that any rational root of the polynomial is a factor of the constant term (1) divided by a factor of the leading coefficient (-1). The possible rational roots are \(\pm 1\).

Let's test these possible roots:

- For \(x = 1\):
\[
-1^3 + 3 \cdot 1^2 - 3 \cdot 1 + 1 = -1 + 3 - 3 + 1 = 0
\]
So, \(x = 1\) is a root.

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.

### 3. **Factoring:**
Using synthetic division to divide \(-x^3 + 3x^2 - 3x + 1\) by \(x - 1\):

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

The quotient is \(-x^2 + 2x - 1\), so we can write the polynomial as:
\[
-x^3 + 3x^2 - 3x + 1 = (x - 1)(-x^2 + 2x - 1)
\]

Next, we factor the quadratic polynomial \(-x^2 + 2x - 1\). We can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = -1\), \(b = 2\), and \(c = -1\):

\[
x = \frac{-2 \pm \sqrt{2^2 - 4(-1)(-1)}}{2(-1)} = \frac{-2 \pm \sqrt{4 - 4}}{-2} = \frac{-2 \pm 0}{-2} = 1
\]

So, the quadratic polynomial \(-x^2 + 2x - 1\) can be factored as \(-(x - 1)^2\).

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

### 4. **Summary:**
The polynomial \(-x^3 + 3x^2 - 3x + 1\) can be factored as \(-(x - 1)^3\). The root of the polynomial is \(x = 1\) with multiplicity 3.