The sum of the fourth powers of the roots of the equation `x^(3)- x^(2) -2x + 2=0` is:

Answer

To find the sum of the fourth powers of the roots of the equation \(x^3 - x^2 - 2x + 2 = 0\), we can follow these steps:

### Step 1: Identify the roots and their relationships
Let \(a\), \(b\), and \(c\) be the roots of the polynomial. By Vieta's formulas, we know:
- The sum of the roots \(a + b + c = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -\frac{-1}{1} = 1\).
- The sum of the products of the roots taken two at a time \(ab + ac + bc = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} = -2\).
- The product of the roots \(abc = -\frac{\text{constant term}}{\text{coefficient of } x^3} = -2\).

### Step 2: Calculate \(a^2 + b^2 + c^2\)
Using the identity:
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) \]
Substituting the values we found:
\[ a^2 + b^2 + c^2 = (1)^2 - 2(-2) = 1 + 4 = 5 \]

### Step 3: Calculate \(a^4 + b^4 + c^4\)
Using the identity:
\[ a^4 + b^4 + c^4 = (a^2 + b^2 + c^2)^2 - 2(a^2b^2 + b^2c^2 + c^2a^2) \]
We first need to find \(a^2b^2 + b^2c^2 + c^2a^2\). This can be expressed as:
\[ a^2b^2 + b^2c^2 + c^2a^2 = (ab + ac + bc)^2 - 2abc(a + b + c) \]
Substituting the known values:
\[ a^2b^2 + b^2c^2 + c^2a^2 = (-2)^2 - 2(-2)(1) = 4 + 4 = 8 \]

Now substituting back into the equation for \(a^4 + b^4 + c^4\):
\[ a^4 + b^4 + c^4 = (5)^2 - 2(8) = 25 - 16 = 9 \]

### Final Answer
Thus, the sum of the fourth powers of the roots is:
\[ \boxed{9} \]