Take b = 2.

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

11 is prime.

b^((n-1)/11)-1 mod n = 6243, which is a unit, inverse 16021.

7 is prime.

b^((n-1)/7)-1 mod n = 7793, which is a unit, inverse 17224.

(7^3 * 11) divides n-1.

(7^3 * 11)^2 > n.

n is prime by Pocklington's theorem.