Is this real number an integer?

Question

Is this real number :

$$\Big(2+\frac{10}{9}\sqrt{3}\Big)^{1/3}+\Big(2-\frac{10}{9}\sqrt{3}\Big)^{1/3}$$

an integer ?

I've tried different factorization, but nothing seems to work.

Answer 1

Let $a = \sqrt[3]{2+\frac{10}{9}\sqrt{3}}$ and $b = \sqrt[3]{2-\frac{10}{9}\sqrt{3}}$. Then:

$a^3+b^3 = (2+\frac{10}{9}\sqrt{3}) + (2-\frac{10}{9}\sqrt{3}) = 4$

$ab = \sqrt[3]{(2+\frac{10}{9}\sqrt{3})(2-\frac{10}{9}\sqrt{3})} = \sqrt[3]{4-\frac{10^2}{9^2}\cdot 3} = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}$.

Now, let $x = a+b$. Then, $x^3 = (a+b)^3 = (a^3+b^3)+3ab(a+b) = 4+2x$.

Therefore, $x$ is a real root of $x^3-2x-4 = 0$. Since $x^3-2x-4 = (x-2)(x^2+2x+2)$, it follows that the only real root of $x^3-2x-4 = 0$ is $x = 2$.