Let \(f(a,b)=a^3+a^2b+ab^2+b^3\).

First, since \(N \le 10^{18}\), when the answer is \(f(a,b)\), then \(a,b \le 10^6\). (This is because, being \(f(10^6,0)=10^{18}\), if at least one of \(a\) and \(b\) is greater than \(10^6\), we have \(f(a,b)>10^{18}\).)

Thus, the objective of this problem is to find for all \(0 \le i \le 10^6\) the minimum non-negative integer \(j\) such that \(f(i,j) \ge N\) and find \(\min f(i,j)\) over all of them.

Here, let \(i_1\) and \(i_2\) be non-negative integers, \(j_1\) be the \(j\) corresponding to \(i=i_1\), and \(j_2\) be the \(j\) corresponding to \(i=i_2\); then it holds that \(j_1 \ge j_2\). (When \(f(i_1,j_1) \ge N\), it holds that \(f(i_2,j_1) \ge N\).)

By this property, the following algorithm is valid.

• First, initialize a variable \(j=10^6\). Let \(X= \infty\) be a variable to store the candidate of the solution.
• For each of \(i = 0\) through \(10^6\), repeat the following.
  • While \(f(i,j) \ge N\) and \(j \ge 0\), repeat the following.
    • Let \(X=\min(X,f(i,j))\), and then decrement \(j\).

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

long long f(long long a,long long b){
  return (a*a*a + a*a*b + a*b*b + b*b*b);
}

int main(){
  long long n;
  cin >> n;
  long long x=8e18;
  long long j=1000000;
  for(long long i=0;i<=1000000;i++){
    while(f(i,j)>=n && j>=0){
      x=min(x,f(i,j));
      j--;
    }
  }
  cout << x << '\n';
  return 0;
}

Here is another implementation equivalent to this algorithm, which is faster by a constant factor. The implementation is essentially that of sliding window technique.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

long long f(long long a,long long b){
  return (a*a*a + a*a*b + a*b*b + b*b*b);
}

int main(){
  long long n;
  cin >> n;
  long long x=8e18;
  long long i=0,j=1000000;
  while(i<=j){
    long long cf=f(i,j);
    if(cf>=n){x=min(x,cf);j--;}
    else{i++;}
  }
  cout << x << '\n';
  return 0;
}