<s>[INST] 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 $x_n=\frac{i}{2}(z_n+\frac{5}{3})$,
 [/INST] Assume $x$ is pure imaginary.
$x=iy$, with $y$ real.
We have:
$$- 2i y^3 +5 i y^2 +3iy +4 i =0$$
or
$$2 y^3 -5  y^2 -3y -4  =0$$
This leads to the solution
$$y \approx 3.17174.$$
The other solutions are
$$y \approx -0.335868 \mp 0.719557 i$$
and the solutions to the original problem are $x=iy.$
There are techniques to solve cubic equations exactly.  I admit that I am not so familiar with them (we could look them up!), so you've got here a numerical solution.
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