Yes, but it’s different:
(a - b) mod p = ((a mod p - b mod p) + p) mod p
(a / b) mod p = ((a mod p) * (b^(-1) mod p)) mod p
Where b^(-1) mod p is the modular inverse of b mod p. For p = prime, b^(-1) mod p = b^(p - 2) mod p.
Edit:
(N*(N^2+5)/6)%P
You don’t need any modular inverses from this. Just simplify the fraction: N or N^2+5 will be divisible by 2 and 3. So divide them and then you have (a*b) mod P.