The given expression is:

\[
\frac{x^3 - 3x^2 + 2x - 1}{x^3 - 1}
\]

We can simplify this step by step.

### Step 1: Factor the denominator
The denominator is \(x^3 - 1\), which is a difference of cubes. We use the formula for factoring the difference of cubes:

\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]

Here, \(x^3 - 1 = (x - 1)(x^2 + x + 1)\).

So, the expression becomes:

\[
\frac{x^3 - 3x^2 + 2x - 1}{(x - 1)(x^2 + x + 1)}
\]

### Step 2: Factor or simplify the numerator
Now, we attempt to factor the numerator \(x^3 - 3x^2 + 2x - 1\). We can use synthetic division or trial division to see if \(x - 1\) is a factor of the numerator as well.

Let's perform synthetic division of \(x^3 - 3x^2 + 2x - 1\) by \(x - 1\):

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

The result of the division is \(x^2 - 2x - 1\) with a remainder of \(-1\), which means that \(x - 1\) is not a factor of the numerator.

### Step 3: Final expression
Since the numerator \(x^3 - 3x^2 + 2x - 1\) cannot be factored further and the remainder isn't zero, the expression cannot be simplified further in terms of canceling terms. The simplified version is:

\[
\frac{x^3 - 3x^2 + 2x - 1}{(x - 1)(x^2 + x + 1)}
\]

This is the most simplified form of the given expression.

Would you like more details or further clarification?

Here are 5 related questions to expand on this:
1. How do you perform synthetic division in polynomial expressions?
2. Can you factor \(x^3 - 3x^2 + 2x - 1\) using another method?
3. What is the difference of cubes and how is it applied?
4. How would you solve for roots of \(x^3 - 1\)?
5. What if the numerator was \(x^3 - 1\), how would the expression change?

**Tip:** Always check if both the numerator and denominator share common factors before simplifying a rational expression.

The query submitted by the user: x^3 - 3x^2 + 2x - 1 / x^3 - 1

Learn how to simplify the polynomial expression (x^3 - 3x^2 + 2x - 1) / (x^3 - 1) using synthetic division and the difference of cubes formula. This problem covers key algebraic concepts including polynomial division and factoring. Suitable for high school students in Grades 10-12.

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