$p(x)$ is the cubic $x^3 - 3x^2 + 5x$. If $h$ is a real root of $p(x) = 1$ and $k$ is a real root of $p(x) = 5$, find $h + k$.
1. Given the cubic polynomial \( p(x) = x^3 - 3x^2 + 5x \), we need to find the sum \( h + k \) where \( h \) is a real root of \( p(x) = 1 \) and \( k \) is a real root of \( p(x) = 5 \).

2. Start by setting up the equations for \( h \) and \( k \):
   \[
   p(h) = 1 \implies h^3 - 3h^2 + 5h = 1
   \]
   \[
   p(k) = 5 \implies k^3 - 3k^2 + 5k = 5
   \]

3. Rearrange the equations:
   \[
   h^3 - 3h^2 + 5h - 1 = 0 \implies h^3 - 3h^2 + 3h - 1 = -2h \implies (h-1)^3 = -2h
   \]
   \[
   k^3 - 3k^2 + 5k - 5 = 0 \implies k^3 - 3k^2 + 3k - 1 = 4 - 2k \implies (k-1)^3 = 4 - 2k
   \]

4. Add the two equations:
   \[
   (h-1)^3 + (k-1)^3 = -2h + (4 - 2k)
   \]
   \[
   (h-1)^3 + (k-1)^3 = 4 - 2(h + k)
   \]

5. Use the identity for the sum of cubes:
   \[
   (h-1)^3 + (k-1)^3 = \big[(h-1) + (k-1)\big]\big[(h-1)^2 - (h-1)(k-1) + (k-1)^2\big]
   \]
   \[
   (h+k-2)\big[(h-1)^2 - (h-1)(k-1) + (k-1)^2\big] = 4 - 2(h+k)
   \]

6. Simplify the equation:
   \[
   (h+k-2)\big[(h-1)^2 - (h-1)(k-1) + (k-1)^2\big] = -2(h+k-2)
   \]

7. Factor out \( h+k-2 \):
   \[
   (h+k-2)\big[(h-1)^2 - (h-1)(k-1) + (k-1)^2 + 2\big] = 0
   \]

8. Since the second factor is strictly greater than 0, we conclude:
   \[
   h+k-2 = 0 \implies h+k = 2
   \]

The final answer is \( \boxed{ h + k = 2 } \)