To have "always one box remaining," the minimum number of boxes must be the Least Common Multiple of 5, 6, and 8, ... PLUS 1.

As products of prime numbers,

5 = 5

6 = 2 * 3

8 = 2 * 2 * 2

The LCM of 5, 6, and 8 is 2*2*2*3*5 = 120. One more is **121**.

Placing 5 boxes in each of 121 cages leaves 1 left over.

Placing 6 boxes in each of 121 cages leaves 1 left over.

Placing 8 boxes in each of 121 cages leaves 1 left over.

That is the situation described !!