$\cos \left( \frac{1}{3}\arccos \frac{37}{64}-\frac{\pi }{3} \right)=\frac{{{\left( -37-3\text{i}\sqrt{303} \right)}^{1/3}}+{{\left( -37+3\text{i}\sqrt{303} \right)}^{1/3}}}{2}$, 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?

  • $\begingroup$ Why do you think it is possible to do this? $\endgroup$ – Thomas Andrews Feb 13 '16 at 5:19
  • $\begingroup$ See casus irreducibilis. $\endgroup$ – Lucian Feb 13 '16 at 5:24
  • $\begingroup$ @ThomasAndrews Just an expection. $\endgroup$ – yaoliding Feb 13 '16 at 5:27

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.

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

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