We can write $(n^3 - 1) = n^3 - 1^3$
Now, applying the formula $$a^3 - b^3 = (a – b)(a^2 + ab + b^2)$$
$$\implies n^3 - 1 = (n - 1)(n^2 + n + 1)$$
We observe that $(n^3 - 1)$ is divisible by $(n - 1)$.
Since $(n^3 - 1)$ is divisible by $(n - 1)$, so, if some $2^k$ divides $(n - 1)$, then that same $2^k$ will divide $(n^3 - 1)$ as $(n - 1)$ is a factor of $(n^3 - 1)$. We don't take into account $(n^2 + n + 1)$ as it is always odd, i.e., for both even and odd $n$. So, $(n^2 + n + 1)$ is not divisible by $2$.
Therefore, $f(n^3 - 1) = f(n - 1)$.
This is true for all $n > 2$.
For $n = 2$, $$(n^3 - 1) = (2^3 - 1) = 8 - 1 = 7$$ which is the only prime number of the form $(n^3 - 1)$. Since in question, they have clearly written that $n > 3$, the answer is $$f(n^3 - 1) = f(n - 1)$$
i.e., Option $(A)$.