Take b = 2.

b^(n-1) mod n = 1.

7 is prime. b^((n-1)/7)-1 mod n = 124, which is a unit, inverse 162.

3 is prime. b^((n-1)/3)-1 mod n = 326, which is a unit, inverse 143.

(3^3 * 7) divides n-1.

(3^3 * 7)^2 > n.

n is prime by Pocklington's theorem.