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.

Maybe you can use the generating function of $x^{\frac 13}$ and show that corresponding terms cancel? Or raise the entire expression to the third power? — Ragnar, Aug 16 '14 at 18:43
Wolfram alpha says it's equal to $2$: http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=(2%2B10%2F9%203%5E(1%2F2))%5E(1%2F3)%2B(2-10%2F9%203%5E(1%2F2))%5E(1%2F3) — Ragnar, Aug 16 '14 at 18:44
Make the ansatz $$(a+b\sqrt{3})^3 = 2 + \frac{10}{9}\sqrt{3}$$ with rational $a,b$. — Daniel Fischer, Aug 16 '14 at 18:53
http://math.stackexchange.com/questions/835955/simple-solving-skanavi-book-exercise-sqrt39-sqrt80-sqrt39-sqrt80 — lab bhattacharjee, Aug 16 '14 at 19:07

score 10 · Answer 1 · answered Aug 16 '14 at 18:58

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 $x = 2$.

1 Answers1