If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, then $p^3+q^3+r^3$ equals

$ \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ \text{none of these} $
Given the polynomial \( x^3 - x^2 + x - 2 = 0 \) with roots \( p, q, \) and \( r \), we need to find the value of \( p^3 + q^3 + r^3 \).

1. **Identify the coefficients:**
   The polynomial is \( x^3 - x^2 + x - 2 = 0 \). The coefficients are:
   \[
   a_3 = 1, \quad a_2 = -1, \quad a_1 = 1, \quad a_0 = -2
   \]

2. **Use Vieta's formulas:**
   Vieta's formulas relate the coefficients of the polynomial to sums and products of its roots:
   \[
   p + q + r = -\frac{a_2}{a_3} = -\frac{-1}{1} = 1
   \]
   \[
   pq + qr + rp = \frac{a_1}{a_3} = \frac{1}{1} = 1
   \]
   \[
   pqr = -\frac{a_0}{a_3} = -\frac{-2}{1} = 2
   \]

3. **Apply Newton's Sums:**
   Newton's sums provide a way to relate the power sums of the roots to the coefficients of the polynomial. Let \( P_k = p^k + q^k + r^k \). Newton's sums state:
   \[
   P_1 - a_2 = 0
   \]
   \[
   P_2 - a_2 P_1 + 2a_1 = 0
   \]
   \[
   P_3 - a_2 P_2 + a_1 P_1 - 3a_0 = 0
   \]

4. **Calculate \( P_1 \):**
   \[
   P_1 - (-1) = 0 \implies P_1 = 1
   \]

5. **Calculate \( P_2 \):**
   \[
   P_2 - (-1)P_1 + 2(1) = 0 \implies P_2 + P_1 + 2 = 0 \implies P_2 + 1 + 2 = 0 \implies P_2 = -1
   \]

6. **Calculate \( P_3 \):**
   \[
   P_3 - (-1)P_2 + 1P_1 - 3(-2) = 0
   \]
   \[
   P_3 + P_2 + P_1 + 6 = 0
   \]
   Substitute \( P_1 = 1 \) and \( P_2 = -1 \):
   \[
   P_3 - 1 + 1 + 6 = 0 \implies P_3 + 6 = 0 \implies P_3 = -6
   \]

Thus, the value of \( p^3 + q^3 + r^3 \) is \(\boxed{-6}\).