Count all triplets whose sum is equal to a perfect cube

Given an array of n integers, count all different triplets whose sum is equal to the perfect cube i.e, for any i, j, k(i < j < k) satisfying the condition that a[i] + a[j] + a[j] = X³ where X is any integer. 3 ? n ? 1000, 1 ? a[i, j, k] ? 5000 

Example: 

Input:
N = 5
2 5 1 20 6
Output:
3
Explanation:
There are only 3 triplets whose total sum is a perfect cube.
Indices Values SUM
0 1 2 2 5 1 8
0 1 3 2 5 20 27
2 3 4 1 20 6 27
Since 8 and 27 are perfect cube of 2 and 3.

Naive approach is to iterate over all the possible numbers by using 3 nested loops and check whether their sum is a perfect cube or not. The approach would be very slow as time complexity can go up to O(n³).

An Efficient approach is to use dynamic programming and basic mathematics. According to the given condition sum of any of three positive integers is not greater than 15000. Therefore there can be only 24(15000^1/3) cubes are possible in the range of 1 to 15000. 
Now instead of iterating all triplets we can do much better by the help of above information. Fixed first two indices i and j such that instead of iterating over all k(j < k ? n), we can iterate over all the 24 possible cubes, and for each one, (let's say P) check how many occurrence of P - (a[i] + a[j]) are in a[j+1, j+2, ... n]. 
But if we compute the number of occurrences of a number say K in a[j+1, j+2, ... n] then this would again be counted as slow approach and would definitely give TLE. So we have to think of a different approach. 
Now here comes to a Dynamic Programming. Since all numbers are smaller than 5000 and n is at most 1000. Hence we can compute a DP array as, 
dp[i][K]:= Number of occurrence of K in A[i, i+1, i+2 ... n] 

Output:

 3

Time complexity: O(N²*24) 
Auxiliary space: O(10⁷)