Is it possible to write $\cos \left( \frac{1}{3}\arccos \frac{37}{64}-\frac{\pi }{3} \right)$ as a radical expression of real number

Question

$\cos \left( \frac{1}{3}\arccos \frac{37}{64}-\frac{\pi }{3} \right)=\frac{{{\left( -37-3\text{i}\sqrt{303} \right)}^{1/3}}+{{\left( -37+3\text{i}\sqrt{303} \right)}^{1/3}}}{2}$, the number inside the cubic root is the complex number, not a real number.

In general for cubic equation $x^3+px+q=0$ (p,q rational number), if $\frac {p^3}{27}+\frac{q^2}4<0$, how to change the form of sulution $\sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} +\sqrt[3]{-{q\over 2} +\sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}$ into an expression as the combination and product of square root, cubic root of rational number, where the number inside the square root must be positive?

Why do you think it is possible to do this? – Thomas Andrews Feb 13 '16 at 05:19 — Thomas Andrews, Feb 13 '16 at 05:19
@ThomasAndrews Just an expection. – Liding Yao Feb 13 '16 at 05:27 — Liding Yao, Feb 13 '16 at 05:27

Tito Piezas III · Accepted Answer · 2016-02-13T16:07:22.817

Unfortunately, it's been known for centuries that it can't be done.

Your cubic belongs to the class $C(x)$ of $n$th degree equations irreducible over the rationals and with all real roots. Not surprisingly, it is called casus irreducibilis.

In general, what you ask can be done for $C(x)$ if and only if $n$ is a power of $2$.
If $n\neq2^m$, like for $n = 3,5,6,$ etc, then you have to go through complex numbers.

Is it possible to write $\cos \left( \frac{1}{3}\arccos \frac{37}{64}-\frac{\pi }{3} \right)$ as a radical expression of real number

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