Proving irrationality of $\sqrt[3]{3}+\sqrt[3]{9}$ [duplicate]

Question

I need to prove $$\sqrt[3]{3}+\sqrt[3]{9}$$ is irrational, I assumed $$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m}{n}$$ I cubed both sides and got $$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m^3-12n^2}{9n^3}$$

I tried setting $$\frac{m^3-12n^2}{9n^3} = \frac{m}{n}$$ but that led me nowhere. so what can I do?

Answer 1

Let $\sqrt[3]3+\sqrt[3]9=r$.

Thus, since for all reals $a$, $b$ and $c$ we have: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain: $$3+9-r^3+9r=0.$$

Now, let $r=\frac{m}{n},$ where $m$ and $n$ are naturals with $gcd=1.$

Thus, $$m^3-9n^2m-12n^3=0,$$ which says that $m$ is divisible by $3$.

Let $m=3m'$, where $m'$ is a natural number.

Thus, $$9m'^3-9n^2m'=4n^3,$$ which says that $n$ is divisible by $3$, which is a contradiction.

Answer 2

Here are two other takes.

Take 1

Let $\alpha = \sqrt[3]{3}+\sqrt[3]{9}$. Then $\alpha^3 = 9 \alpha + 12$.

By the rational root theorem, if $\alpha$ is rational, then $\alpha$ is an integer.

Now $1 < \sqrt[3]{3} < 2 $ and $2 < \sqrt[3]{9} < 3 $, and so $3 < \alpha < 5$.

Since $x=4$ is not a root of $x^3 = 9 x + 12$, $\alpha$ is not an integer and so $\alpha$ is irrational.

Take 2

We have $\alpha = \beta+\beta^2$, where $\beta=\sqrt[3]{3}$.

If $\alpha$ were rational, then $\beta$ would be a root of a quadratic polynomial with rational coefficients. However, the polynomial with rational coefficients with least degree having $\beta$ as a root is $x^3-3$.

Answer 3

Let $a=\sqrt[3]{3}$ and $b=\sqrt[3]{9}$, and suppose that $a+b=r$ is rational. Then $$r^3 =(a+b)^3 = a^3+3ab(a+b)+b^3 = 12+9(a+b)=12+9r$$

So $r$ is a solution of equation $x^3-9x-12=0$ which has no rational root.