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.