The cubic $x^3=px+q$ with $p,q\in \mathbb{R}$ has the formula
$$x=\sqrt[3]{\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}{3}\right)^3}}+\sqrt[3]{\frac{q}{2}-\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}{3}\right)^3}}$$
When $\left(\frac{q}{2}\right)^2-\left(\frac{p}{3}\right)^3<0$ we have the cube roots of two complex numbers which are conjugates, so the answer is real. If $z=\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}{3}\right)^3}$ then we want $2\textrm{Re}\left(z^{1/3}\right)$. How does one get this last result in real form without solving another cubic?
**Edit. **
To clarify, this is easily done analytically: once we have $z$ we can get its polar form $re^{i\theta}$ then $z^{1/3}=r^{1/3}e^{i\theta/3}$ and then $2\textrm{Re}\left(z^{1/3}\right)=2r^{1/3}\cos\left(\theta/3\right)$. However, if one wants $\cos\left(\theta/3\right)$ in terms of $tan^{-1}\left(\theta\right)$, one finds another cubic. Is there an algebraic way out? i.e. using arithmetic operations and $n^{\textrm{th}}$ -roots of reals only?
Best Answer
In general, the real and imaginary parts of solutions to polynomial equations whose degree is $3$ or more can not be expressed as a finite sequence of arithmetic operations and $n$th root extractions ($n$ being a positive integer) over the real numbers.