Take b = 2.

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

89 is prime.

b^((n-1)/89)-1 mod n = 5873, which is a unit, inverse 7292.

3 is prime.

b^((n-1)/3)-1 mod n = 3085, which is a unit, inverse 2175.

(3^3 * 89) divides n-1.

(3^3 * 89)^2 > n.

n is prime by Pocklington's theorem.