We can use the fact that $p$, $q$, and $r$ are the roots of the equation to rewrite it as \[(x-p)(x-q)(x-r) = 0.\] Expanding, we get \[x^3 - (p+q+r)x^2 + (pq+pr+qr)x - pqr = 0.\] Since the coefficients of the equation are the same as the coefficients of $x^3 - x^2 + x - 2 = 0$, we can equate the corresponding coefficients to get \[\begin{aligned} p+q+r &= 1, \\ pq+pr+qr &= 1, \\ pqr &= 2. \end{aligned}\] We want to find $p^3 + q^3 + r^3$. Notice that \[(p+q+r)(p^2+q^2+r^2-pq-qr-pr) = p^3+q^3+r^3 - 3pqr.\] Using the equations we obtained earlier, we can rewrite this as \[1(p^2+q^2+r^2-1) = p^3+q^3+r^3 - 6.\] Since $p^2+q^2+r^2 = (p+q+r)^2 - 2(pq+pr+qr) = 1^2 - 2(1) = -1$, we have \[1(-1-1) = p^3+q^3+r^3 - 6.\] Solving for $p^3+q^3+r^3$, we find that $p^3+q^3+r^3 = \boxed{4}$.The answer is: 4
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
If $p$, $q$, and $r$ are the roots of the equation $x^3 - x^2 + x - 2 = 0$, what is the value of $p^3 + q^3 + r^3$?

### Answer:
