Base case: For \(n = 1\), \(5^{3 \times 1} + 32^1 - 33 = 5^3 + 32 - 33 = 124\), which is divisible by 31.

Inductive step: Assume \(5^{3k} + 32^k - 33\) is divisible by 31 for some positive integer \(k\).

Consider \(5^{3(k+1)} + 32^{k+1} - 33\):

\(5^{3(k+1)} + 32^{k+1} - 33 = 5^{3k+3} + 32^{k+1} - 33\)

\(= (124 + 1)5^{3k} + (31 + 1)32^k - 33\)

\(= 124 \times 5^{3k} + 31 \times 32^k + 5^{3k} + 32^k - 33\)

Since \(124 \times 5^{3k} + 31 \times 32^k\) is divisible by 31, \(5^{3(k+1)} + 32^{k+1} - 33\) is divisible by 31.

Thus, by induction, the statement is true for every positive integer \(n\).