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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?

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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.

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In general, if $z=a+bi$ you can take its cube roots by converting to polar form. Write $z=re^{i\theta}$ with $r=\sqrt{a^2+b^2}, \theta=\pm \arctan \frac ba$ where you choose the sign to get the correct quadrant. Then $z^{1/3}=r^{1/3}e^{i\theta/3+2k\pi/3}$ where $k$ is any integer.

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  • $\begingroup$ Sorry this is not what I meant, I should've explained: the solution you propose would then give $x=2r^{1/3}\cos\left(\theta /3+2\pi k\right)$. If you can calculate cos and tan, then this works. However, to try to find $\cos\left(\tan^{-1} \left(x\right)/3\right)$ algebraically, one ends up solving a cubic. So I meant an algebraic way out. $\endgroup$ Oct 11, 2019 at 15:27
  • $\begingroup$ @JoshuaTilley no you don’t have to if you invoke trig identities. $\endgroup$ Oct 11, 2019 at 15:34
  • $\begingroup$ @ElenKhachatryan, the one-third angle formula needed gives rise to a new cubic. $\endgroup$ Oct 11, 2019 at 15:36

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