Define f(n) = n^6+24n^3+192
The simplest solutions would be when f(n) = n^6. Trying that implies 24n^3+192 = 0, which makes n=-2 and f(n)=64 one solution.
Lets make a compound inequality (n^2-1)^3 < f(n) < (n^2+1)^3. The only perfect cube between (n^2-1)^3 and (n^2+1)^3 is n^6. If there are any integer solutions when f(n) != n^6 then n must fail the compound inequality.
(n^2-1)^3 < f(n) is false for -7.999<=n<=-2.171 and f(n) < (n^2+1)^3 is false for -1.831<=n<=7.999. Combining these then limits integer n to the range -7 to 7. No new solutions are in this range so set of integer n for which f(n) is a perfect cube is a single value n=-2.
Solution
