is there away to give $= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}$

Question

question is there away to give

$$= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}$$

in just reals?

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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

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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 $$ \begin{aligned} x&=\sqrt[3]{\frac{10}{2} + \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\ + \sqrt[3]{\frac{10}{2} - \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\ \\&= \sqrt[3]{5 + \sqrt{25 -125}} + \sqrt[3]{5- \sqrt{25 -125}} \\ &= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i} \end{aligned} $$

Answer 1

When there are $3$ real roots, Cardano-Tartaglia's formulæ can't be applied, and we need a trigonometric solution:

Setting $x=A\cos \theta\;$ ($A>0,\; 0\le \theta <\pi$), the equation becomes $$A^3\cos^3\theta -15A \cos\theta=10.$$ We'll try to write the l.h.s. as $\;B\cos 3\theta=10\;$ to obtain an easy-to-solve trigonometric equation. For this , we need to satisfy the proportionality equation $$\frac{A^3}4=\frac{15A}{3}\iff A^2=20\iff A=2\sqrt 5.$$ So the equation becomes $$10\sqrt 5\cos 3\theta=10\iff\cos 3\theta=\frac{\sqrt 5}5.$$ The general solutions are $$3\theta\equiv \pm\arccos\Bigl(\frac{\sqrt 5}5\Bigr)\mod 2\pi\iff \theta\equiv\pm\frac13\arccos\Bigl(\frac{\sqrt 5}5\Bigr)\mod\frac{2\pi}{3}. $$ One checks the three solutions in the interval $[0,\pi]$ are $$ \theta= \frac13\arccos\Bigl(\frac{\sqrt 5}5\Bigr),\quad \theta=\frac13\arccos\Bigl(\frac{\sqrt 5}5\Bigr)+\frac{2\pi}3,\theta =-\frac13\arccos\Bigl(\frac{\sqrt 5}5\Bigr)+\frac{4\pi}3.$$

Answer 2

No there isnt. This is a well known phenomenon with a latin name, Casus Irreducibilus.

If a third degree equation with real coefficients is irreducible, and has all real solutions then these solutions cannot be expressed as roots of real numbers.

Answer 3

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 $$\theta =\cos^{-1} (\frac {\sqrt 5}{5})$$

Answer 4

Since $$ (5+10i)^{1/3}=\sqrt5\,e^{i\arctan(2)/3} $$ and $$ (5-10i)^{1/3}=\sqrt5\,e^{-i\arctan(2)/3} $$ Thus, $$ \bbox[5px,border:2px solid #C0A000]{(5+10i)^{1/3}+(5-10i)^{1/3}=2\sqrt5\cos\left(\frac{\arctan(2)}3\right)} $$

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Another Approach to the Original Question

Let $\alpha=2\sqrt5$. Then, using $\cos(3x)=4\cos^3(x)-3\cos(x)$, we get $$ \begin{align} 10 &=\frac{\alpha^3}4\cos(3x)\\ &=\alpha^3\cos^3(x)-\frac34\alpha^3\cos(x)\\[5pt] &=\underbrace{\alpha^3\cos^3(x)}_{z^3}-15\,\underbrace{\ \alpha\cos(x)\ }_z \end{align} $$ Thus, with $\cos(3x)=\frac{40}{\alpha^3}=\frac1{\sqrt5}$, we get that $z=2\sqrt5\cos(x)$ is a root. This gives $3$ real roots $$ z_1=2\sqrt5\cos\left(\frac13\arccos\left(\frac1{\sqrt5}\right)\right) $$ $$ z_2=2\sqrt5\cos\left(\frac{2\pi}3+\frac13\arccos\left(\frac1{\sqrt5}\right)\right) $$ $$ z_3=2\sqrt5\cos\left(\frac{4\pi}3+\frac13\arccos\left(\frac1{\sqrt5}\right)\right) $$ Noting that $\arccos\left(\frac1{\sqrt5}\right)=\arctan(2)$, we get that $z_1=(5+10i)^{1/3}+(5-10i)^{1/3}$.