<s>[INST] is there away to give $= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}$ question is there away to give 
$$= \sqrt[3]{5 + 10i } 
 + \sqrt[3]{5- 10i}$$
in just reals?

I am thinkting maybe use some clever algebra raise it to like 3 and magically imaginary terms dissapear. Or expressed the complex numbers in polar form but idk

Background (might delete if this is not really the question)
Let $$ m(x)=x^3-15x-10 $$
Find the roots of $m(x)$ using the cubic formula and show that they are all real 
Using Howell complex approach 
Lets change variables $$ \begin{aligned}
&z^3-15z-10
\\=&z^3+0z^2-15z-10 
\end{aligned}$$
so, general form is denoted as 
$$ z^3+a_2z^2+a_1z+a_0=0$$
so $$\begin{aligned}
a_2&=0
\\a_1&=-15
\\a_0&=-10
\end{aligned}$$
we get by substituting that $z=x-a_2/3$. $a_2=0$ its already depressed cubic but so theses steps are unnecesary for this one problem but will use them for other problems
$$ x^3+bx+c$$
where $$\begin{aligned}
b&=a_1-\frac{a_2^2}{3}
                       =-15-\frac{0^2}{3}=-15
\\c&=\frac{-a_1a_2}{3}+\frac{2}{27}a_2^3+a_0
                        =\frac{-(-15)(0)}{3}+\frac{2}{27}(0)^3+-10=-10
\end{aligned} $$
our depressed cubic is of $$  x^3-15x-10 $$
Ferro-Tartaglia Formula 
$$ x=\sqrt[3]{\frac{-c}{2} + \sqrt{\frac{c^2}{4} +\frac{b^3}{27}}}\ 
+ \sqrt[3]{\frac{-c}{2} - \sqrt{\frac{c^2}{4} -\frac{b^3}{27}}}\ $$
in our case $$ 
\begin{aligned}
x&=\sqrt[3]{\frac{10}{2} + \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\ 
+ \sqrt[3]{\frac{10}{2} - \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\ 
 \\&= \sqrt[3]{5 + \sqrt{25 -125}} 
 + \sqrt[3]{5- \sqrt{25 -125}}
\\ &=  \sqrt[3]{5 + 10i } 
 + \sqrt[3]{5- 10i}
\end{aligned}
$$
 [/INST] Note that $$ \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}=
Z+ \bar Z =2\operatorname{Re}(Z)$$
Thus it is a real number. 
In order to find the real part of Z, we may write it in polar form and get $$\operatorname{Re}(Z) = \sqrt 5 \cos(\theta /3)$$ where $$\theta =\cos^{-1} (\frac {\sqrt 5}{5})$$
</s>