i have some ideas here.. we can use root=r=p/q p and q are integers and p/q is irriducible. there are two cases. p and q are both odd, or p(or q) is even. we have another two cases, b and c are both even; or c or b is even. goodluck

i found three cases, in two of them the root is irrational b and c and p(or q) are even. than ap+bpq+cpq+dq=even+even+even+odd= odd=0 which is not true because 0 is not odd. b and c are even, p and q are odd ap+bpq+cpq+dq=odd+even+even+odd=even=0 which is true b OR c are even, p and q are odd ap+bpq+cpq+dq=odd +even+odd+odd=odd=0 which is not true because 0 is not odd.

FTSOC let all the roots of $ax^3 + bx^2 + cx + d$ be rational. Since, $ad$ is odd $\implies a,d$ are odd. Hence, $ax^3 + bx^2 + cx + d = ((2a_0+1)x - (2a_1 + 1))((2b_0+1)x - (2b_1 + 1))((2c_0+1)x - (2c_1 + 1))$ By this we get that $c = (2a_0+1)(2b_1 + 1)(2c_1 + 1) + (2b_0+1)(2c_1 + 1)(2a_1 + 1) + (2c_0+1)(2a_1 + 1)(2b_1 + 1)$ = odd + odd + odd = odd Similarly $b = -[(2a_0+1)(2b_0 + 1)(2c_1 + 1) - (2b_0+1)(2c_0 + 1)(2a_1 + 1) - (2c_0+1)(2a_0 + 1)(2b_1 + 1)]$ = odd + odd + odd = odd Hence, $bc =$ odd $\times$ odd = odd $\implies$ Contradiction!