Answer: For every positive integer \(n\), \(f(n)=2^{3n}+8^{n-1}\) is divisible by \(9\).

We first simplify the expression.

Since \(2^{3n}=(2^3)^n=8^n\), we have

\(f(n)=8^n+8^{n-1}=8^{n-1}(8+1)=9\cdot 8^{n-1}.\)

So \(f(n)\) is a multiple of \(9\) for every positive integer \(n\).

If we present this as an induction argument, the result can be shown as follows.

Base case: When \(n=1\),

\(f(1)=2^3+8^0=8+1=9,\)

which is divisible by \(9\).

Inductive step: Assume that for some positive integer \(k\), \(f(k)\) is divisible by \(9\). We must show that \(f(k+1)\) is also divisible by \(9\).

Now

\(f(k)=2^{3k}+8^{k-1}=8^k+8^{k-1}=9\cdot 8^{k-1},\)

and

\(f(k+1)=2^{3k+3}+8^k=8^{k+1}+8^k=9\cdot 8^k.\)

Hence \(f(k+1)\) is divisible by \(9\).

Therefore, by mathematical induction, \(f(n)\) is divisible by \(9\) for every positive integer \(n\).