There are only \(O(N^\frac{1}{3})\) cube numbers less than or equal to \(N\). Thus, one can bruteforce over all cube numbers \(N\) and check if its decimal representation is a palindrome, in order to solve this problem.

Sample code

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

bool check(ll n) {
    string s = to_string(n);
    string t = s;
    reverse(t.begin(), t.end());
    return s == t;
}

int main() {
    ll n;
    cin >> n;
    ll ans = 0;
    for (ll i = 1; i * i * i <= n; i++) if (check(i * i * i)) ans = i * i * i;
    cout << ans << '\n';
}