Let $1=d_1<d_2<d_3<\dots<d_k=n$ be the divisors of $n$. Find all values of $n$ such that $n=d_2^2+d_3^3$.
1. **Assume \( n \) is odd.**
   - We know that \( d_2 \) is the smallest divisor of \( n \) other than 1, so it must be a prime number.
   - If \( n \) is odd, then all its divisors \( d_1, d_2, \ldots, d_k \) must also be odd.
   - However, \( d_2^2 + d_3^3 \) would be even if both \( d_2 \) and \( d_3 \) are odd, which contradicts the assumption that \( n \) is odd.
   - Therefore, \( n \) must be even.

2. **Assume \( n \) is even.**
   - Since \( n \) is even, \( d_2 \) must be the smallest prime divisor of \( n \), and \( d_2 \) must be 2 (the smallest prime number).
   - Given \( n = d_2^2 + d_3^3 \), we substitute \( d_2 = 2 \):
     \[
     n = 2^2 + d_3^3 = 4 + d_3^3
     \]
   - Since \( d_2 \) is a prime number, \( d_2 \) must divide \( d_3^3 \). Given \( d_2 = 2 \), it implies \( 2 \mid d_3^3 \), so \( d_3 \) must be even.
   - The smallest even number greater than 2 is 4. Therefore, \( d_3 = 4 \).

3. **Verify the value of \( n \):**
   - Substitute \( d_3 = 4 \) into the equation:
     \[
     n = 4 + 4^3 = 4 + 64 = 68
     \]
   - Check the divisors of 68: \( 1, 2, 4, 17, 34, 68 \).
   - The divisors are \( 1 = d_1, 2 = d_2, 4 = d_3, 17 = d_4, 34 = d_5, 68 = d_6 \).
   - The condition \( n = d_2^2 + d_3^3 \) holds true:
     \[
     68 = 2^2 + 4^3 = 4 + 64 = 68
     \]

Therefore, the only value of \( n \) that satisfies the given condition is \( n = 68 \).

The final answer is \( \boxed{68} \)