Find $n$ .

We have $A=n^3-2n^2+2n-4=(n-2)(n^2+2)$ . Since for all $n, n-2<n^2+2$ , therefore for $A$ to be prime, $n-2$ must be $1$ . For $n-2=1$ or $n=3, n^2+2=11$ , a prime number. Hence the only possible value of $n$ is $\boxed 3$ .

If $f(n)=n^3-2n^2+2n-4$ , then $f(3)=11$ , which is prime.

Also, since $f(n)=(n-2)(n^2+2)$ , $f(n)$ is composite for all $n>3$ (and it's easily verified that $f(n)$ is not positive for $n<3$ ) so $n=3$ is in fact the only integer that makes $f(n)$ prime.