The number of real roots of `x^(8) - x^(5) + x^(2) - x + 1 = 0` is

Explanation

To determine the number of real roots of the equation \( x^8 - x^5 + x^2 - x + 1 = 0 \), we will analyze the function step by step.

### Step 1: Define the function
Let \( f(x) = x^8 - x^5 + x^2 - x + 1 \).

### Step 2: Analyze the function
We will examine the behavior of the function \( f(x) \) to determine if it can equal zero for any real \( x \).

### Step 3: Check the limits as \( x \) approaches infinity and negative infinity
- As \( x \to \infty \), the leading term \( x^8 \) dominates, so \( f(x) \to \infty \).
- As \( x \to -\infty \), the leading term \( x^8 \) (since it is even) also dominates, so \( f(x) \to \infty \).

### Step 4: Check for critical points
To find critical points, we will compute the derivative \( f'(x) \):
\[ f'(x) = 8x^7 - 5x^4 + 2x - 1 \]
We need to find where \( f'(x) = 0 \) to identify potential local maxima or minima.

### Step 5: Analyze the derivative
Since \( f'(x) \) is a polynomial of degree 7, it can have up to 7 real roots. However, we will check the nature of \( f'(x) \) to understand the behavior of \( f(x) \).

### Step 6: Evaluate \( f(x) \) at specific points
- Calculate \( f(0) \):
\[ f(0) = 0^8 - 0^5 + 0^2 - 0 + 1 = 1 \]
- Calculate \( f(1) \):
\[ f(1) = 1^8 - 1^5 + 1^2 - 1 + 1 = 1 - 1 + 1 - 1 + 1 = 1 \]
- Calculate \( f(-1) \):
\[ f(-1) = (-1)^8 - (-1)^5 + (-1)^2 - (-1) + 1 = 1 + 1 + 1 + 1 + 1 = 5 \]

### Step 7: Check the sign of \( f(x) \)
From the evaluations:
- \( f(0) = 1 \) (positive)
- \( f(1) = 1 \) (positive)
- \( f(-1) = 5 \) (positive)

### Step 8: Conclude the number of real roots
Since \( f(x) \) approaches infinity as \( x \) approaches both positive and negative infinity, and since \( f(x) \) is positive at the evaluated points, we can conclude that \( f(x) \) does not cross the x-axis. Therefore, the equation \( f(x) = 0 \) has no real roots.

### Final Answer
The number of real roots of the equation \( x^8 - x^5 + x^2 - x + 1 = 0 \) is **0**.