We know the square roots of a complex number can be written as
where when otherwise
Thus we have real algebraic expressions for the real and imaginary parts of the square root, without trigonometric functions involved.
Does this still the case for cubic roots? I tried with no success. Thought that's not possible but cannot prove my guess
Ok, I'll show you in general you can't express and in terms of radicals.
From the two equations we can get (solve for in equation 1 and substitute into equation 2).
Now square both sides:
Simplify:
Since there is no formula that can solve for a polynomial of degree greater than in terms of radicals, we see that this equation is unsolvable in general.
Applying this method to yields the same results.
Let such that
Then when we have as a solution.
If , then and we have
hence
Let , then we have
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 solvable in real field?
Or how to disprove the equation is solvable in real field?