I'd like to share my doubt on cubic equation.
Step 1: $ax^3+bx^2+cx+d=0$,
Step 2: We can substitute $x=y-\frac b {3a}$ to get $y^3+py+q=0$ where $p,q$ are something.
Step 3: By Vieta's substitution, we substitute $z-\frac p {3z}=y$ and get $z^6+qz^3-\frac {p^3}{27}=0$
Step 4: $z^3=-\frac q 2\pm\sqrt{(\frac q 2)^2+(\frac p 3)^3}$
Step 5: Substitute $A=-\frac q 2+\sqrt{(\frac q 2)^2+(\frac p 3)^3},B=-\frac q 2-\sqrt{(\frac q 2)^2+(\frac p 3)^3}$, we get $z=\sqrt[3]A,\sqrt[3]A\omega,\sqrt[3]A\omega^2,\sqrt[3]B,\sqrt[3]B\omega$ or $\sqrt[3]B\omega^2$ where $\omega=\frac {\sqrt 3i-1}2$.
Step 6: $AB=-\frac {p^3}{27},y=z-\frac p {3z}$,therefore $y=z+\frac {\sqrt[3]{AB}} z$.
Here, I don't understand. Why needn't we consider $\sqrt[3]{AB}=\frac p 3\omega$ or $\frac p 3\omega^2$.
Thank you.