roots - Proof that $sqrt[m]{a} + sqrt[n]{b}$ is irrational

Is there a way to prove that $\sqrt[m]{a} + \sqrt[n]{b}$ ($\sqrt[m]{a}$ and $\sqrt[n]{b}$ are irrational); $a, b, m, n \in \mathbb{N}$; $m, n \neq 2$; is irrational without using the theorem mentioned in Sum of irrational numbers, a basic algebra problem?

If one of $m$ or $n$ is $2$, then a polynomial with integer coefficients can be easily constructed, and rational root theorem (http://en.wikipedia.org/wiki/Rational_root_theorem) can be used to prove that it's irrational. For example, if $x = \sqrt{2} + \sqrt[3]{3}$:

$$\begin{align} (x - \sqrt{2})^3 = x^3 - 3x^2\sqrt{2} + 6x - 2\sqrt{2} & = 3 \\ \implies x^3 + 6x - 3 &= \sqrt{2}(3x^2 + 2) \\ \implies x^6 + 12x^4 - 6x^3 + 36x^2 - 36x + 9 & = 2(9x^4 + 12x^2 + 4) \\ \implies x^6 - 6x^4 - 6x^3 + 12x^2 - 36x + 5 & = 0 \end{align}$$

By evaluating the polynomial for $\pm1$ and $\pm5$, it can be verified that $x$ is irrational. However, if neither of $m$ or $n$ is $2$, then constructing a polynomial with integer coefficients seems impossible (if not very tedious). Let's say $x = \sqrt[3]{2} + \sqrt[4]{3}$. Is there any way to prove that this is irrational without using the above-mentioned theorem?