show that the equation $3z^3+(2-3ai)z^2+(6+2bi)z+4=0$has exactly one real root

Question

show that :

the equation $3z^3+(2-3ai)z^2+(6+2bi)z+4=0$ (where both $a$ and $b$ are real numbers) has exactly one real root.

let $x_{1},x_{2},x_{3}$ be root,and $$x_{1}x_{2}x_{3}=-\frac{4}{3},x_{1}+x_{2}+x_{3}=\dfrac{3ai-2}{3},x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}=\dfrac{2bi+6}{3}$$

Answer 1

If we had exactly 3 real roots, then the sum of all roots would be real, but it is not.

If we had exactly 2 real roots, then the product of all roots would not be real, but it is.