O! These rads are fun

Let $\color{#D61F06}{A=\sqrt[3]{5+2\sqrt{13}}}$ , $\color{#3D99F6}{B=\sqrt[3]{5-2\sqrt{13}}}$ and $\color{#20A900}S=\color{#D61F06}A+\color{#3D99F6}B=\color{#D61F06}{\sqrt[3]{5+2\sqrt{13}}}+\color{#3D99F6}{\sqrt[3]{5-2\sqrt{13}}}$

Now,

$\begin{aligned}\color{#20A900}S^3=&\ (A+B)^3\\=&\ A^3+3A^2B+3AB^2+B^3\\=&\ A^3+B^3+3AB(\color{#D61F06}A+\color{#3D99F6}B)\\=&\ \color{#302B94}{A^3+B^3}+\color{magenta}{3AB}\color{#20A900}S\Rightarrow(*)\end{aligned}$

Thus,

$\begin{aligned}\color{#302B94}{A^3+B^3}=&\ \left(\color{#D61F06}{\sqrt[3]{5+2\sqrt{13}}}\right)^3+\left(\color{#3D99F6}{\sqrt[3]{5-2\sqrt{13}}}\right)^3\\=&\ 5+2\sqrt{13}+5-2\sqrt{13}\\=&\ \color{#302B94}{10}\end{aligned}$

$\begin{aligned}\color{magenta}{3AB}=&\ 3\left(\color{#D61F06}{\sqrt[3]{5+2\sqrt{13}}}\right)\left(\color{#3D99F6}{\sqrt[3]{5-2\sqrt{13}}}\right)\\=&\ 3\sqrt[3]{(5+2\sqrt{13})(5-2\sqrt{13})}\\=&\ 3\sqrt[3]{25-52}\\=&\ 3\sqrt[3]{-27}\\=&\ 3(-3)\\=&\ \color{magenta}{-9}\end{aligned}$

Instead $\color{#302B94}{A^3+B^3}$ with $\color{#302B94}{10}$ and $\color{magenta}{3AB}$ with $\color{magenta}{-9}$ in equation $(*)$

$\begin{aligned}\color{#20A900}S^3=&\ \color{#302B94}{10}+(\color{magenta}{-9})\color{#20A900}S\\\color{#20A900}S^3+9\color{#20A900}S-10=&\ 0\\(\color{#20A900}S-1)(\color{#20A900}S^2+\color{#20A900}S+10)=&\ 0\\\color{#20A900}{S=}&\ \color{#20A900}1\end{aligned}$

From equation

$\begin{aligned}\color{#20A900}S^2+\color{#20A900}S+10=&\ 0\\\Delta=&\ 1^2-4(1)(10)=-39\end{aligned}$

$\Delta<0$ . Thus, this equation has no real solution.

Hence, $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}=\boxed1$

$Let\;A=\sqrt[3]{5+2\sqrt{13}}\\ B=\sqrt[3]{5-2\sqrt{13}}\\ Then \; x=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}=A+B\rightarrow (1)$ Multiply both sides of equation $(1)$ by $(A^2-AB+B^2)=((A+B)^2-3AB)$ to get $x\times ((A+B)^2-3AB)=A^3+B^3$ .Now: $AB=\sqrt[3]{5+2\sqrt{13}}\times\sqrt[3]{5-2\sqrt{13}}=\sqrt[3]{25-52}=\sqrt[3]{-27}=-3\\ A^3+B^3=5+2\sqrt{13}+5-2\sqrt{13}=10\\ \implies x\times ((A+B)^2-3AB)=A^3+B^3=x \times (x^2+9)=10\rightarrow (2)$ By inspection, $x=1$ satisfies (2).Dividing $x^3+9x-10$ by $x-1$ gives $x^2+x+10\Rightarrow \Delta=1^2-4(1)(10)=-39< 0$ .As $x\in \mathbb{R}$ so $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}=\boxed{1}$

Let $\sqrt [3]{5+2\sqrt {13}}+\sqrt [3]{5-2\sqrt{13}}$ be $x$ .Now cube both sides to get: $5+2\sqrt {13}+5-2\sqrt {13}+3\sqrt [3]{(5+2\sqrt {13})(5-2\sqrt {13})}×(\sqrt [3]{5+2\sqrt {13}}+\sqrt[3]{5-2\sqrt {13}})=x^{3} \\ \Rightarrow{10+3x\sqrt [3]{-27}=x^{3 }}$ ( because $\sqrt [3]{5+2\sqrt {13}}+\sqrt [3]{5-2\sqrt{13}}=x$ .) $\Rightarrow{10-9x=x^{3}} \Rightarrow{x^{3}+9x-10=0}$ Now trying with $x=1$ satisfies the above equation above. Note that as it is a cubic equation,there are 2 more solutions of $x$ .Let's obtain 'em as well. So $(x-1)(x^{2}+x+10=0)\Rightarrow{x^{2}+x+10=0}$ clearly $\Delta<0$ ,so the equation has no real solutions for $x$ . So the value of $\sqrt [3]{5+2\sqrt {13}}+\sqrt [3]{5-2\sqrt{13}}$ is $\boxed{1}$ . $\ddot\smile$

After solving, and seeing my solution was the same as most, I decided to come up with a different solution, so I could share something different. Perhaps someone could try to properly justify the reasoning of this approach using the theory of fields, because I honestly don't know the proper justification (although it is correct). Anyway... Let $\sqrt[3]{5+2\sqrt{13}}=a+b\sqrt{13}$ and $\sqrt[3]{5-2\sqrt{13}}=a-b\sqrt{13}$ . Multiplying the two equations results in $-3=a^2-13b^2$ . Then take the equation $\sqrt[3]{5+2\sqrt{13}}=a+b\sqrt{13}$ and cube both sides: $5+2\sqrt{13}=a^3+3a^2b\sqrt{13}+39ab^2+13\sqrt{13}b^3$ . Equating the terms that do not contain $\sqrt{13}$ , I get $5=a^3+39ab^2$ . If I take the equation $-3=a^2-13b^2$ , multiply it by $3a$ , and then add it to $5=a^3+39ab^2$ , I get $4a^3+9a-5=0$ . Using the rational zeros theorem to test for roots, one can determine $a=\frac{1}{2}$ , with the other two roots being imaginary. Since the sum is equal to $2a$ (the $b\sqrt{13}$ and $-b\sqrt{13}$ cancel), the answer is $2a=2(\frac{1}{2})=1$ . (And in case anyone was wondering, even though the polynomial I got looks different than the ones in the other solutions, if I let $x=2a$ , and replace $a$ with $\frac{x}{2}$ , then the polynomial will be the same as in the other solutions.)

5^1/3 + 2 . 13^1/2 + 3^1/2 . 13^1/2 = 1.