Cube root of complex number without trigonometric functions

Is there a general equation for a cube root of a complex number that does not exploit De Moivre's Theorem or in any way use trigonometric functions?

For example, a square root of a complex number $x$ is $$\sqrt{\frac{|x|+\operatorname{Re}(x)}{2}}+i\sqrt{\frac{|x|-\operatorname{Re}(x)}{2}}.$$ Is there a similar equation for a cube root of $x$?

By introducing $a$ and $b$ such that $(a+ib)^3=x$, we can then expand and obtain two equations in $a$ and $b$ by equating the real and imaginary parts on each side.

However, the resulting equations are cubic and I don't know any method to find the roots of a cubic equation without having to take the cube root of a complex number, which is the problem I want to solve in the first place.

• @BrianTung: No; trisection is impossible by straightedge and compass for a different (and much simpler) reason: each such construction will produce a coordinate in a quadratic extension (of the previous field) if at all, and hence every such constructible point lies in an extension of $\mathbb{Q}$ of degree a power of $2$, and by the tower law must have minimal polynomial with degree also a power of $2$. $\cos(20^\circ)$ has a cubic minimal polynomial over $\mathbb{Q}$, and hence cannot be constructible in this way. But the neusis construction can do general trisection. – user21820 Nov 27 '16 at 13:59
• @BrianTung: Anyway I meant the reason for the impossibilities are not related, and they are related only through the fact that casus irreducibilis is equivalent to irreducible rational cubic with 3 real roots, which is in turn related to the constructibility of cosines (via irreducibility) since the cubic $( 4x^3 - 3x - \cos(t) )$ has 3 real roots (via intermediate value theorem; try $x \in \{-1,-\frac12,0,1\}$). There is no connection with the original definition of casus irreducibilis in terms of radicals. – user21820 Nov 27 '16 at 14:13