Prove that $5^{1/3}-3^{1/4}$ is irrational

Question

Prove that $5^{1/3}-3^{1/4}$ is irrational.

Let $x=5^{1/3}-3^{1/4}$. I started by:

\begin{align} (x+3^{1/4})^3&=5\\ x^3+3^{3/4}+3x\cdot3^{1/4}5^{1/3}&=5\\ 3^{3/4}+3x\cdot3^{1/4}5^{1/3}&=5-x^3 \tag{1} \end{align}

Now I wished to prove by contradiction that - assuming $x$ is rational - LHS of (1) is irrational whereas RHS is rational, so the equation cannot be satisfied. However, the LHS is a sum of irrational terms, and we know that the sum of irrational terms is not always irrational.

I obviously don't wish to apply power four on (1) . Is there any other shorter method?

Answer 1

~An answer involving elementary field theory

If it were rational then we would have $5^{1/3} = q + 3^{1/4}$ where $q$ is a rational number. This would then imply that the field $K = \mathbb{Q}(3^{1/4})$ would contain the field $\mathbb{Q}(5^{1/3})$ which obviously cannot be the case as $[\mathbb{Q}(3^{1/4}):\mathbb{Q}] = 4$ and $[\mathbb{Q}(5^{1/3}):\mathbb{Q}] = 3$ and certainly $3$ does not divide $4$.

Answer 2

The minimal polynomial of $5^{1/3} - 3^{1/4}$ is (according to Wolfram Alpha) $$P(x) = x^{12} - 20 x^9 - 9 x^8 + 150 x^6 - 720 x^5 + 27 x^4 - 500 x^3 - 2250 x^2 - 540 x + 598$$ According to the rational roots theorem every rational root of $P$ is a divisor of $598$. In particular every rational root of $P$ is an integer.

You can show that $0 < 5^{1/3} - 3^{1/4} < 1$ so it is not an integer, hence not rational.