Let, $R, C$ be functions of $z$, and everything else except $R,C$ are rational number and the below equation holds-
$$\sqrt [3] {R + \sqrt {Q^3 + R^2} } + \sqrt [3] {R - \sqrt {Q^3 + R^2} } - \dfrac b {3 a}=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$$
In other words,
$$\sqrt [3] {R(z) + \sqrt {Q^3 + R(z)^2} } + \sqrt [3] {R(z) - \sqrt {Q^3 + R(z)^2} } - \dfrac b {3 a}=\frac{-B\pm\sqrt{B^2-4AC(z)}}{2A}$$ $$\implies f(z)=\sqrt [3] {R(z) + \sqrt {Q^3 + R(z)^2} } + \sqrt [3] {R(z) - \sqrt {Q^3 + R(z)^2} } - \dfrac b {3 a}-\frac{-B\pm\sqrt{B^2-4AC(z)}}{2A}=0$$
Can we say $z$ is a rational number?
I have tried by simplifying the equation but it gets complicated, is there any other way to prove that $z$ is a rational/irrational number number?
Edit:
$C(z)=(z+w_1)^2, R(z)= \frac{z^2-z}{2}+w_2$ where $w_1, w_2$ are known rational number.