An algebra problem by vansh gupta

Let $x = a + b$ where $a = \sqrt[3]{5 + 2\sqrt{13}}$ and $b = \sqrt[3]{5 - 2\sqrt{13}}$ . Then

$x^{3} = (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} = a^{3} + b^{3} + 3ab(a + b) = a^{3} + b^{3} + 3abx$ .

Now $a^{3} + b^{3} = (5 + 2\sqrt{13}) + (5 - 2\sqrt{13}) = 10$ , and

$ab = \sqrt[3]{(5 + 2\sqrt{13})(5 - 2\sqrt{13})} = \sqrt[3]{25 - 4 \times 13} = \sqrt[3]{-27} = -3$ . so

$x^{3} = 10 + 3x \times (-3) = 10 - 9x \Longrightarrow x^{3} + 9x - 10 = (x - 1)(x^{2} + x + 10) = 0$ .

As the roots of $x^{2} + x + 10$ are complex, we can conclude that the real solution for $x$ is $\boxed{1}$ .

Let $x = \sqrt[3]{5+2\sqrt{13}} + \sqrt[3]{5-2\sqrt{13}}$ , then:

$\begin{aligned} x^3 & = \left( \sqrt[3]{5+2\sqrt{13}} + \sqrt[3]{5-2\sqrt{13}}\right)^3 \\ & = 5+2\sqrt{13} + 3 \sqrt[3]{(5+2\sqrt{13})^2(5-2\sqrt{13})} + 3 \sqrt[3]{(5+2\sqrt{13})(5-2\sqrt{13})^2} + 5-2\sqrt{13} \\ & = 10 + 3 \sqrt[3]{(5+2\sqrt{13})(5^2-(2\sqrt{13})^2)} + 3 \sqrt[3]{(5^2-(2\sqrt{13})^2)(5-2\sqrt{13})} \\ & = 10 + 3 \sqrt[3]{(5+2\sqrt{13})(-27)} + 3 \sqrt[3]{(-27)(5-2\sqrt{13})} \\ & = 10 - 9 \sqrt[3]{5+2\sqrt{13}} - 9 \sqrt[3]{5-2\sqrt{13}} \\ & = 10 - 9x \end{aligned}$

Therefore, we have:

$\begin{aligned} x^3 + 9x - 10 & = 0 \\ (x-1)(x^2+x+10) & = 0 \\ \implies x & = \boxed{1} \end{aligned}$

Since $x^2+x+10 = 0$ has no real root.