I think one of them is always divisible by $5$. So one of them must be equal to $5$. Check the cases.

In Detail : $n^3-4n^2+3n-35=n(n-3)(n-1)-7\cdot 5$ $\implies 5\nmid n(n-3)(n-1)$ $\implies n\equiv 2,4\pmod{5}$ Now we observe that : 1) $n^2+4n+8=(5k+2)^2+4(5k+2)+8\equiv 2^2+4\cdot 2+3\equiv 0\pmod{5}$ 2) $n^2+4n+8=(5z+4)^2+4(5z+4)+8\equiv 16+16+8\equiv 0\pmod{5}$ Therefor $n^2+4n+8$ is always divisible by $5$ which implies it is equal to $5$ $\implies n^2+4n+8=(n+3)(n+1)+5=5$ $\implies (n+3)(n+1)=0\implies n=-3,-1$ Now we just have to plug it in $|n^3-4n^2+3n-35|$ and check whether it is also a prime or not.

@above you forgot when $5\mid n$ so $n^3-4n^2+3n=n (n^2-4n+3)=40$ so $n=5$ and $n^2+4n+8=53$ thus $\boxed {n=5~\text {is a solution}}$.