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, and find this root.
I've dealt with quadratics in this form but never with cubics. I tried searching up the cubic formula but I think it would take way too long to be practical in an exam situation..
If $z$ is real, you can split the equation in real/imaginary parts:
$$3z^3+2z^2+6z+4=0\\-3az^2+2bz=0.$$
From the second equation, you can exclude $z=0$ and you get $z=\dfrac{2b}{3a}$. Plugging in the first equation, the numerator is
$$8b^3+8ab^2+36ba^2+36a^3=0.$$
By factoring, there must be a solution $a+b=0$, so that $\color{green}{z=-\dfrac23}$. The other factor is $8b^2+36a^2$, which is irreducible.
Remains to show that the initial cubic equation can have no other real root. For this, we divide the polynomial by $3z+2$, and
$$3z^3+(2-3ai)z^2+(6+2bi)z+4=(3z+2)\left(z^2-iaz+\frac{6+2i(a+b)}{3}\right)-\frac{4i}3(a+b).$$
With $a+b=0$, this simplifies the equation to
$$z^2-iaz+2=0.$$
Assuming $z$ real once again,
$$z^2+2=0\\az=0$$ which is not possible.