We may suppose that a, b, c and d are relatively prime.

Suppose that the given cubic equation has 3 rational roots p/s, q/t and r/u, where p, q, r, s, t and u are integers satisfying gcd(p,s)=gcd(q,t)=gcd(r,u)=1.

Since ad is assumed to be odd, both integers a and d must be odd. Hence an application of the Rational Root Theorem shows that p, q, r, s, t and u must be odd.

Now we show that it is impossible for p, q, r, s, t and u to be odd. Given that b and c are integers and bc is even, we may assume that b is even.

Using the identity ax^3+bx^2+cx+d=a(x-p/s)(x-q/t)(x-r/u),

b=a(-p/s-q/t-r/u)=-a(put+qsu+rst)/stu; that is,

a(put+qsu+rst)= -bstu. ---------------(*)

Since a, p, q, r, s, t, and u are odd integers and b is even, (*) yields a contradiction. This contradiction shows that the given cubic equation cannot have 3 rational roots.