What is the solution set of the equation `x^(3) - x^(2) + x - 1 gt 0` ?

Options

A. `x lt -1`
B. `-1 lt x lt 1`
C. `x gt 1`
D. `x gt 0`

Answer

To solve the inequality \( x^3 - x^2 + x - 1 > 0 \), we will follow these steps:

### Step 1: Factor the polynomial
We start by factoring the polynomial \( x^3 - x^2 + x - 1 \). We can group the terms:

\[ x^3 - x^2 + x - 1 = (x^3 - x^2) + (x - 1) \]

Now, factor out common terms from each group:

\[ = x^2(x - 1) + 1(x - 1) \]

This gives us:

\[ = (x - 1)(x^2 + 1) \]

### Step 2: Analyze the factors
Now we have the inequality:

\[ (x - 1)(x^2 + 1) > 0 \]

Next, we analyze each factor:

1. **\( x - 1 \)**: This factor is zero when \( x = 1 \). It is positive when \( x > 1 \) and negative when \( x < 1 \).
2. **\( x^2 + 1 \)**: This factor is always positive for all real \( x \) because the square of any real number is non-negative, and adding 1 ensures it is always greater than zero.

### Step 3: Determine the intervals
Since \( x^2 + 1 > 0 \) for all \( x \), the sign of the product \( (x - 1)(x^2 + 1) \) is determined solely by \( (x - 1) \).

- For \( x < 1 \): \( (x - 1) < 0 \) → The product is negative.
- For \( x = 1 \): The product is zero.
- For \( x > 1 \): \( (x - 1) > 0 \) → The product is positive.

### Step 4: Write the solution set
We want the product to be greater than zero:

\[ (x - 1)(x^2 + 1) > 0 \implies x > 1 \]

Thus, the solution set is:

\[ \boxed{(1, \infty)} \]