The sum of the real roots of the equation `x^(5)-5x^(4)+9x^(3)-9x^(2)+5x-1=0` is

Answer

To find the sum of the real roots of the polynomial equation

\[ x^5 - 5x^4 + 9x^3 - 9x^2 + 5x - 1 = 0, \]

we will follow these steps:

### Step 1: Identify a Rational Root
We can use the Rational Root Theorem to test for possible rational roots. Testing \(x = 1\):

\[ 1^5 - 5(1^4) + 9(1^3) - 9(1^2) + 5(1) - 1 = 1 - 5 + 9 - 9 + 5 - 1 = 0. \]

Since \(x = 1\) is a root, we can factor the polynomial by \(x - 1\).

### Step 2: Perform Polynomial Long Division
We will divide the polynomial by \(x - 1\):

\[ x^5 - 5x^4 + 9x^3 - 9x^2 + 5x - 1 \div (x - 1). \]

After performing the division, we find:

\[ x^5 - 5x^4 + 9x^3 - 9x^2 + 5x - 1 = (x - 1)(x^4 - 4x^3 + 5x^2 - 4x + 1). \]

### Step 3: Solve the Quartic Equation
Now we need to solve the quartic equation:

\[ x^4 - 4x^3 + 5x^2 - 4x + 1 = 0. \]

### Step 4: Use Substitution
Let \(y = x + \frac{1}{x}\). Then, we can express \(x^2 + \frac{1}{x^2}\) in terms of \(y\):

\[ x^2 + \frac{1}{x^2} = y^2 - 2. \]

Substituting into the quartic equation, we can rewrite it in terms of \(y\).

### Step 5: Solve for \(y\)
We find the values of \(y\) from the transformed equation. This will lead us to two quadratic equations:

1. \(x^2 - x + 1 = 0\) (which has complex roots).
2. \(x^2 - 3x + 1 = 0\) (which has real roots).

### Step 6: Find the Real Roots
Using the quadratic formula on \(x^2 - 3x + 1 = 0\):

\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}. \]

### Step 7: Sum of the Real Roots
The real roots are \(x = 1\) from the first factor and \(x = \frac{3 + \sqrt{5}}{2}\) and \(x = \frac{3 - \sqrt{5}}{2}\) from the second factor.

Calculating the sum:

\[ \text{Sum} = 1 + \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = 1 + \frac{3 + \sqrt{5} + 3 - \sqrt{5}}{2} = 1 + \frac{6}{2} = 1 + 3 = 4. \]

### Final Answer
The sum of the real roots of the equation is:

\[ \boxed{4}. \]