radicals - How to prove $sqrt3 + sqrt[3]{2}$ is a irrational number?

This is an exercise from R. Courant's book: How to prove $\sqrt3 + \sqrt[3]{2}$ is a irrational number?

The solution is to construct a equation to prove, but is there any other method to prove this, like by contradiction?

Answer

Developing the hint by Winther. Assume it is rational $r=\sqrt{3}+\sqrt[3]{2}$. So we have

$$2=(r-\sqrt{3})^3=r^3-3r^2\sqrt{3}+9r-3\sqrt{3}$$

Regrouping and dividing by $3r^2+3\neq 0$ we get

$$\sqrt{3}={r^3+9r-2\over 3r^2+3}$$

This means $\sqrt{3}$ is rational. Contradiction