Let $k$ be a positive integer. Find the maximum value of \[a^{3k-1}b+b^{3k-1}c+c^{3k-1}a+k^2a^kb^kc^k,\] where $a$, $b$, $c$ are non-negative reals such that $a+b+c=3k$.
1. **Assume WLOG**: Without loss of generality, we can assume that \(a \geq b \geq c\). This is because the expression is symmetric in \(a, b, c\), and we can always reorder the variables to satisfy this condition.

2. **Define the function**: Let 
   \[
   f(a, b, c) = a^{3k-1}b + b^{3k-1}c + c^{3k-1}a + k^2 a^k b^k c^k
   \]
   We aim to find the maximum value of \(f(a, b, c)\) under the constraint \(a + b + c = 3k\).

3. **Reduction to two variables**: We want to show that \(f(a, b, c) \leq f(a+c, b, 0)\). This means we need to prove:
   \[
   (a+c)^{3k-1}b \geq a^{3k-1}b + b^{3k-1}c + c^{3k-1}a + k^2 a^k b^k c^k \quad (1)
   \]

4. **Expand the LHS**: When we expand \((a+c)^{3k-1}b\), we get:
   \[
   (a+c)^{3k-1}b = \sum_{i=0}^{3k-1} \binom{3k-1}{i} a^i c^{3k-1-i} b
   \]
   The significant terms in this expansion are:
   \[
   a^{3k-1}b + (3k-1)a^{3k-2}bc + \binom{3k-1}{2} a^{3k-3}bc^2 + \cdots + c^{3k-1}b
   \]

5. **Compare terms**: For \(k=1\), the inequality follows from the first three and the last summands. For \(k \geq 2\), the inequality follows from the first four summands because \(\binom{3k-1}{2} \geq k^2\).

6. **Conclusion**: Hence, the maximum is reached when \(c=0\). Therefore, we have:
   \[
   f(a, b, c) \leq (3k-1)^{3k-1}
   \]

7. **Equality conditions**: Equality holds for \(k=1\) when \((a, b, c) = (1, 1, 1)\) or \((2, 1, 0)\), and for \(k \geq 2\) when \((a, b, c) = (3k-1, 1, 0)\) with cyclic permutations.

\(\blacksquare\)

The final answer is \( \boxed{ (3k-1)^{3k-1} } \)