How to prove that $\sqrt[3] 2 + \sqrt[3] 4$ is irrational? [duplicate]

So while doing all sorts of proving and disproving statements regarding irrational numbers, I ran into this one and it quite stumped me:

Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.

I tried all the usual suspects like playing with $\sqrt[3]{2} + \sqrt[3]{4} = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ , but got nowhere.

I also figured maybe I should play with it this way:

$2^\frac{1}{3} + 4^\frac{1}{3}=2^\frac{1}{3} + (2^2)^\frac{1}{3}=2^\frac{1}{3} + 2^\frac{2}{3}=2^\frac{1}{3} + 2^\frac{1}{3}\times 2^\frac{1}{3}=2^\frac{1}{3}(1+2^\frac{1}{3})$

But there I got stumped again, because while $1+2^\frac{1}{3}$ is irrational, nothing promises me that $2^\frac{1}{3} \times (1+2^\frac{1}{3})$ is irrational, and I feel like trying to go further down this road is moot.

So what am I missing (other than sleep and food)? What route should I take to prove this? Thanks in advance!

Note

$$1 + \sqrt[3]{2} + \sqrt[3]{4} = \frac{(\sqrt[3]{2})^3 - 1}{\sqrt[3]{2} - 1} = \frac{1}{\sqrt[3]{2} - 1}.$$

So if $\sqrt[3]{2} + \sqrt[3]{4}$ is rational, then $1/(\sqrt[3]{2} - 1)$ is rational, which implies $\sqrt[3]{2} - 1$ is rational. Then $\sqrt[3]{2}$ is rational, a contradiction.

$y = \sqrt[3]{2} + \sqrt[3]{4}$ is a root of the equation $y^3 - 6y - 6 = 0$.

(To see this, let $x = \sqrt[3]{2}$ and $y = \sqrt[3]{2}+\sqrt[3]{4}=x+x^2$. Then $y^3 = x^3 + 3x^4 + 3x^5 + x^6 = 2 + 6x + 6x^2 + 4 = 6 + 6y$.)

By the rational root theorem, we know that any rational roots of $y^3 - 6y - 6 = 0$ would have to be in the set $\{\pm1,\pm2,\pm3,\pm6\}$; we can quickly confirm that none of these is in fact a root, meaning that the equation does not have any rational roots.

Therefore, $y$ must be irrational.

An easy approach: If $p=\sqrt[3]{2}+\sqrt[3]{4}$ is rational:

$$p^2 = \sqrt[3]{4}+2\cdot 2+ 2\sqrt[3]{2} = p+4+\sqrt[3]{2}.$$

So $\sqrt[3]{2}=p^2-p-4$ would be rational.

An alternative, more general approach.

Claim: If $a,b$ are integers that are not perfect cubes, and $a\neq -b$, then $\sqrt[3]{a}+\sqrt[3]b$ is irrational.

Proof:

Assume $\sqrt[3]{a}+\sqrt[3]{b}$ is rational. Cube it and get:

$$(\sqrt[3]{a}+\sqrt[3]{b})^3 = a+ 3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) + b$$

Now, since $a,b$ are rational and $\sqrt[3]{a}+\sqrt[3]{b}$ is non-zero and rational, this means that $\sqrt[3]{ab}$ is rational.

Letting $p=\sqrt[3]{a}+\sqrt[3]{b}$ and $q=\sqrt[3]{ab}$, this means that

$$(x-\sqrt[3]{a})(x-\sqrt[3]{b}) = x^2-px+q$$ is a rational polynomial. It shares at least one root with $x^3-b$, But the GCD of these two polynomials has to be a rational polynomial, so the GCD cannot be linear (since it would be $x-\sqrt[3]b$, which is not a rational polynomial.)

This means that $x^2-px+q$ has to divide $x^3-b$. That means that $\sqrt[3]{a}$ is a root of $x^3-b$, which means that $a=b$. But there are no repeated roots of $x^3-b$, which yields a contradiction.

Corollary: If $a,b$ are rationals such that $a\neq -b$ and $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are irrational, then $\sqrt[3]{a}+\sqrt[3]{b}$ is irrational.

Proof: Rationalize the denominators and revert to the above theorem for integers.

Here is another approach which generalises to many similar examples, and which involves no complicated simplifications of surds. (In fact, no easy simplifications of surds either.)

First, $\sqrt[3]2$ is an algebraic integer, that is, it is a root of $$x^3-2$$ which is a monic polynomial (leading coefficient $1$) with integer coefficients. Similarly, $\sqrt[3]4$ is an algebraic integer.

It is known that the sum of two algebraic integers is an algebraic integer. Thus, $x=\sqrt[3]2+\sqrt[3]4$ is an algebraic integer.

It is also known that if an algebraic integer is rational, then it is a rational integer - that is, an ordinary integer, $0,1,-1,2,-2$ etc. However, it is not hard to find the estimates $$1<\sqrt[3]2<\frac43 ,\quad \frac43<\sqrt[3]4<\frac53\ ;$$ so $\frac73<x<3$, hence $x$ is not an integer and must be irrational.