$\sqrt[31]{12} +\sqrt[12]{31}$ is irrational

Prove that $\sqrt[31]{12} +\sqrt[12]{31}$ is irrational.

I would assume that $\sqrt[31]{12} +\sqrt[12]{31}$ is rational and try to find a contradiction.

However, I don't know where to start. Can someone give me a tip on how to approach this problem?

Let $\mathbb{Q}(\alpha)$ denote the smallest field containing $\mathbb{Q}$ and $\alpha$.

The theory of field extensions tells us that $\mathbb{Q}(\sqrt[31]{12})$ has degree $31$ over $\mathbb{Q}$, $\mathbb{Q}(\sqrt[12]{31})$ has degree $12$ over $\mathbb{Q}$, and, because $(31,12)=1$, we have $\mathbb{Q}(\sqrt[31]{12})\cap\mathbb{Q}(\sqrt[12]{31}) = \mathbb{Q}$.

If $\sqrt[31]{12} +\sqrt[12]{31}$ were a rational number, we would have $\sqrt[31]{12} \in \mathbb{Q}(\sqrt[31]{12})\cap\mathbb{Q}(\sqrt[12]{31}) = \mathbb{Q}$. But $\sqrt[31]{12}$ is not rational, contradiction.

Here is a simpler variant of Slade's answer:

If $\sqrt[31]{12} +\sqrt[12]{31}$ were a rational number, we would have $\sqrt[31]{12} \in \mathbb{Q}(\sqrt[12]{31})$ and so $\mathbb{Q}(\sqrt[31]{12}) \subseteq \mathbb{Q}(\sqrt[12]{31})$.

But $\mathbb{Q}(\sqrt[31]{12})$ has dimension $31$ over $\mathbb{Q}$ and so cannot be a subspace of $\mathbb{Q}(\sqrt[12]{31})$, which has dimension $12$.