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**elim** Let $\displaystyle (a,b),(x,y) \in \mathbf{R}^2$ such that $\displaystyle a+ib=(x+iy)^3 = x^3-3xy^2+i(3x^2y-y^3)$

Then when $\displaystyle a=0$ we have $\displaystyle (x,y)=(0,-\sqrt[3]{b})$ as a solution.

If $\displaystyle a \neq 0$, then $\displaystyle x \neq 0$ and we have

$\displaystyle b^2 = (3x^2y-y^3)^2=\left(\frac{x^3-a}{3x}\right)\left(3x^2-\frac{x^3-a}{3x}\right)^2$ hence

$\displaystyle 64x^9-48ax^6-(15a^2+27b^2)x^3-a^3=0$

Let $\displaystyle x^3=z+a/4, \quad p=-\frac{27}{64}(a^2+b^2),\quad q=-\frac{27a}{256}(a^2+b^2)$, then we have

$\displaystyle z^3+pz+q = 0$ and so z can be expressed in terms of radicals and so can x and y.

But the problem is that the radicals themselves(if using Cardano's method or known methods for me) are involving complex numbers and we get cyclic situations!

Seems we can express the problem as this:

Does real cubic equation $\displaystyle z^3+pz+q = 0$ solvable in real field?

Or how to disprove the equation is solvable in real field?