A bruteforce approach can solve this problem easily

---

include<bits/stdc++.h>

using namespace std;

int main(){

int a[]={1,-2,3,-4,5};

int ans=0;

for(int i=0;i<5;i++){

    for(int j=0;j<5;j++){

        for(int k=0;k<5;k++){

            if(i+j+k==6){

                ans+=a[i]*a[j]*a[k];

            }

        }

    }

}

cout<<ans;

}

The unique products of powers that matter in this problem are those of the form that add up to the 6th power. Namely, $\\ \\ (0 , 2 , 4) \\ (0 , 3 , 3) \\ (1 , 1 , 4) \\ (1 , 2 , 3) \\ (2 , 2 , 2) \\$

Each product repeats depending on the possible permutations.

These are called permutations with repetition.

Then, only 5 products are to be calculated.

$0, 2, 4 \rightarrow$ the coefficients product is $(1 * 3 * 5) = 15,$ which is repeated 6 times $15 *6 = 90 \\ 0, 3, 3 \rightarrow (1 * -4 * -4) = 16, \qquad 16 * 3 = 48 \\ \\ 1, 1, 4 \rightarrow (-2 * -2 * 5) = 20, \qquad 20 * 3 = 60 \\ \\ 1, 2, 3 \rightarrow (-2 * 3 * -4) = 24, \qquad 24*6 = 144 \\ \\ 2, 2, 2 \rightarrow (3 * 3 * 3) = 27, \qquad 27 * 1 = 27 \\$

So, the coefficient we look for is

$90 + 48 + 60 + 144 + 27 = 369.$

Notice that this approach is valid for a harder problem of this type.