Solving $2x^3-5ix^2+3x+4i=0$, need $(21x_1-1)^{-2}+(21x_2-1)^{-2}+(21x_3-1)^{-2}$ I have this polynomial from Problemas selectos (Lumbreras editors):
$$2x^3-5ix^2+3x+4i=0$$
I make $x=it$. then:
$$2(it)^3-5i(it)^2+3it+4i=0$$
$$i(-2it^3+5it^2+3it+4)=0$$
$$2t^3-5t^2-3t-4=0$$
$$8t^3-20t^2-12t-16=0$$
$$(2t)^3-3(2t)^2(5/3)+(50/3)t-125/27-(14/3)t-307/27=0$$
$$(2t-5/3)^3-(86/3)t-307/27=0$$
If $2t-5/3=z$, then:
$$z^3-(43/3)z-952/27=0$$
Where $p=-\frac{43}{3}$ and $q=-\frac{952}{27}$
When $A=-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$ and $B=-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$
We have $z_1=\sqrt[3]{A}+\sqrt[3]{B}$, $z_2=-\frac{\sqrt[3]{A}+\sqrt[3]{B}}{2}+i\sqrt{3}\frac{\sqrt[3]{A}-\sqrt[3]{B}}{2}$, $z_3=-\frac{\sqrt[3]{A}+\sqrt[3]{B}}{2}-i\sqrt{3}\frac{\sqrt[3]{A}-\sqrt[3]{B}}{2}$
There for $x_n=\frac{i}{2}(z_n+\frac{5}{3})$,

Starting from @mjw's answer, you face a cubic equation with only one real root in $y$  since $\Delta=-5447$.
Using the hyperbolic method for this root, we then have
$$y_1=\frac{5}{6}+\frac{\sqrt{43}}{3}  \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\frac{476}{43 \sqrt{43}}\right)\right)=3.17174\cdots$$
Now, deflate the cubic and get the quadratic equation
$y^2+Ay+B=0$ where
$$A=y_1-\frac{5}{2}\qquad \text{and} \qquad B=\frac 2 {y_1}$$
Numerically, they are $y_{2,3}=-0.335868\pm 0.719557\, i$  as already given by @mjw.
