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

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?

• Please use descriptive titles. You have 150 characters to use, and you're more than welcomed to use $\rm\LaTeX$ as well. "Proving the irrationality of a number" is a meaningless title. Same goes to your other post from earlier. Commented Jan 28, 2019 at 18:43
• Closely related: math.stackexchange.com/questions/1542708 Commented Jan 29, 2019 at 9:06
• It may be that this exercise was intended to reinforce specific material covered in your textbook or other course materials, but your Readers will not be aware of that context unless you share it. It is certainly a problem that is amenable to various attacks, so knowing what you are studying will help Readers respond in a way you find useful. Commented Jan 29, 2019 at 17:18

## 3 Answers

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.

• What did you set as a, b, and c? Commented Jan 28, 2019 at 9:07
• @Guysudai1 They are any real numbers. Commented Jan 28, 2019 at 9:07
• @Guysudai1 Michael seems to have chosen $a=\sqrt[3]{3}$, $b=\sqrt[3]{9}$, and $c=-r$. Then $a+b+c=0$ and one can study the left-hand side. Commented Jan 28, 2019 at 13:43

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 $$x^3-3$$.

• It is fine what $\beta$ is a root of $x^3 - 3$ but how do you know that it is of least degree polynomial with rational coefficients. Are you using some standard result? Commented Feb 4, 2019 at 7:26
• @prashantsharma, $x^3 - 3$ is a cubic with no rational roots and so is irreducible.
– lhf
Commented Feb 4, 2019 at 7:58
• @Ihf Means are you suggesting that any equation with no rational roots is irreducible? Commented Feb 5, 2019 at 9:24
• @prashantsharma, not any equation, but cubics, yes.
– lhf
Commented Feb 5, 2019 at 10:12
• @Ihf Can you please supply a proof? Commented Feb 5, 2019 at 10:34

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.