Two cubes?

If Both n + 3 and $n^{2}$ + 3 are cubes,the their product must also be a cube. so $(n+3)({ n }^{ 2 }+3)={ n }^{ 3 }+3{ n }^{ 2 }+{ 3n }+{ 9 }={ (n+1) }^{ 3 }+8$ This can be a perfect cube only if it is 8 more than another perfect cube. The only pairs of perfect cubes that differ by 8 are (−8, 0) and (0, 8). So we must have $n$ = -1, -3. For neither of these solutions is $n^{2}$ + 3 a perfect cube.

Therefore, if n is an integer, $n$ + 3 and $n^{2}$ + 3 cannot both be perfect cubes.