Let $Z^3=a+bi$, $b\not=0$
Is it possible for such a complex number to have real roots for $Z=\sqrt[3]{a+bi}$? I did some calculations and found out that it is not but I have a problem:
The equation $x^3-15x+4=0$ has three real roots, however when I put the coefficients into my cubic formula, which always gives the correct answer except for these cases, it does not give me any real solutions because it has an imaginary number, obtained from $\sqrt{-}$ inside the cube root. One of its real roots is $2+\sqrt{3}$, but what I obtain is $2+\sqrt{3} -i2.22 \times 10^{-16}$. Here, the imaginary number is very small and when I replace $x$ with this in equation, it gives me $0$, which means that this value is correct, even if the solutions for the equation are real.
This is where the problem is in the formula: For $x^3+px+q=0$,
$$\sqrt{\frac{27\times q^2+4\times p^3}{108}}$$
As it is seen here, when $p=-15$ and $q=4$, inside of the square root (which is actually inside of a cube root) becomes negative and makes the inside of the cube root complex number. This results in $3$ complex roots having imaginary coefficients other than $0$. Even if it is very close, as the imaginary part is $-i2.22 \times 10^{-16}$, it is not exactly $0$ where it must be.
The problem is probably about my formula even if I and my friends cannot see, but I would like to learn if such equations are to be treated differently etc.