What is the sum of cubes of real roots of the equation $x^3-2x^2-x+1=0$?

$ 
\textbf{(A)}\ -6
\qquad\textbf{(B)}\ 2
\qquad\textbf{(C)}\ 8
\qquad\textbf{(D)}\ 11
\qquad\textbf{(E)}\ \text{None of above}
$
1. Given the polynomial equation \( f(x) = x^3 - 2x^2 - x + 1 = 0 \), we need to find the sum of the cubes of its real roots.
2. First, we determine the number of real roots by evaluating the polynomial at several points:
   \[
   f(-1) = (-1)^3 - 2(-1)^2 - (-1) + 1 = -1 - 2 + 1 + 1 = -1
   \]
   \[
   f(0) = 0^3 - 2(0)^2 - 0 + 1 = 1
   \]
   \[
   f(1) = 1^3 - 2(1)^2 - 1 + 1 = 1 - 2 - 1 + 1 = -1
   \]
   \[
   f(10) = 10^3 - 2(10)^2 - 10 + 1 = 1000 - 200 - 10 + 1 = 791
   \]
   Since \( f(x) \) changes signs between \(-1\) and \(0\), \(0\) and \(1\), and \(1\) and \(10\), by the Intermediate Value Theorem, there are three real roots.

3. Let the roots of the polynomial be \( x_1, x_2, x_3 \). By Vieta's formulas, we know:
   \[
   x_1 + x_2 + x_3 = 2
   \]
   \[
   x_1 x_2 + x_2 x_3 + x_3 x_1 = -1
   \]
   \[
   x_1 x_2 x_3 = -1
   \]

4. We need to find \( x_1^3 + x_2^3 + x_3^3 \). Using the identity for the sum of cubes:
   \[
   x_1^3 + x_2^3 + x_3^3 = (x_1 + x_2 + x_3)(x_1^2 + x_2^2 + x_3^2 - x_1 x_2 - x_2 x_3 - x_3 x_1) + 3x_1 x_2 x_3
   \]

5. First, we find \( x_1^2 + x_2^2 + x_3^2 \) using the square of the sum of the roots:
   \[
   (x_1 + x_2 + x_3)^2 = x_1^2 + x_2^2 + x_3^2 + 2(x_1 x_2 + x_2 x_3 + x_3 x_1)
   \]
   \[
   2^2 = x_1^2 + x_2^2 + x_3^2 + 2(-1)
   \]
   \[
   4 = x_1^2 + x_2^2 + x_3^2 - 2
   \]
   \[
   x_1^2 + x_2^2 + x_3^2 = 6
   \]

6. Substitute back into the sum of cubes identity:
   \[
   x_1^3 + x_2^3 + x_3^3 = (x_1 + x_2 + x_3)(x_1^2 + x_2^2 + x_3^2 - x_1 x_2 - x_2 x_3 - x_3 x_1) + 3x_1 x_2 x_3
   \]
   \[
   x_1^3 + x_2^3 + x_3^3 = 2(6 - (-1)) + 3(-1)
   \]
   \[
   x_1^3 + x_2^3 + x_3^3 = 2(7) - 3
   \]
   \[
   x_1^3 + x_2^3 + x_3^3 = 14 - 3
   \]
   \[
   x_1^3 + x_2^3 + x_3^3 = 11
   \]

The final answer is \(\boxed{11}\).